## NCERT Solutions for Class 12 Maths Chapter 13 - Probability

The NCERT Solutions for Class 12 Maths Chapter 13 Probability are provided here in PDF format to help students in their board and other competitive exams. Probability is defined as the possibility of the occurrence of an event. In this chapter, students learn about the concept of Probability in detail. Solving all the questions of this chapter given in the NCERT Solutions for Class 12 Math Chapter 13 Probability is very easy as the subject experts at Vedantu ensure that the solutions are given in the simplest form. These solutions cover all the exercise questions included in the book and are prepared as per the guidelines issued by the CBSE board. Students can download these NCERT Solutions for Class 12 Maths Chapter 13 through the PDF link below.

### Topics Covered In Class 12 Maths Chapter 13

Introduction To Probability

Conditional Probability

Properties of Conditional Probability

Multiplication Theorem on Probability

Independent Events

Bayes Theorem

Partition of Sample Space

Theorem of Total Probability

Random Variables and its Probability Distribution

Probability Distribution of Random Variable

Means of Random Variables

Variance of Random Variables

Binomial Distribution

Miscellaneous Exercise

Refer to Probability Class 12 NCERT Solutions available on Vedantu for detailed understanding on probability, which is an important topic in Class 12 Mathematics. Students will learn the basic principles of probability and get a strong base of Chapter 13 Class 12 Maths when they refer to our solutions. Our well-designed NCERT Solutions for Class 12 Maths Chapter 13 have been prepared by knowledgeable teachers and the subject matter experts who have tried to explain this vast concept in as simple terms as possible.

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## Download PDF of NCERT Solutions Class 12 Maths Chapter 13 Probability

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### Exercise-13.1

1. Given that E and F are events such that $P\left( E \right)=0.6$, and $P\left( F \right)=0.3 $\[P\left( E\bigcap F \right)=0.2\], find P(E|F) and P(F|E).

Ans: It is given in the question that $P\left( E \right)=0.6$ = 0.6, $P\left( F \right)=0.3$, and $P\left( E\bigcap F \right)=0.2$

Now P(E|F) is given by

\[\text{ }P\left( E|F \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}\]

$\Rightarrow P\left( E|F \right)=\frac{0.2}{0.3}$

Hence we found that

$P\left( E|F \right)=\frac{2}{3}$

With similar idea and the same formula we can proceed to find P(F|E) as shown

\[P\left( F|E \right)=\frac{P\left( F\bigcap E \right)}{P\left( E \right)}\]

Now we know that\[P\left( F\bigcap E \right)\] and $P\left( E\bigcap F \right)$ is same

$\therefore P\left( F|E \right)=\frac{0.2}{0.6}$

Hence we found that

$P\left( F|E \right)=\frac{1}{3}$

Thus $P\left( E|F \right)=\frac{2}{3}$ and $P\left( F|E \right)=\frac{1}{3}$

2. Compute P(A|B), if $P\left( B \right)=0.5$ and $P\left( A\bigcap B \right)=0.32$

Ans: Given in the question $P\left( B \right)=0.5$ and $P\left( A\bigcap B \right)=0.32$

So to find P(A|B we use the formula

\[P\left( A|B \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}\]

\[P\left( A|B \right)=\frac{0.32}{0.5}\]

$\frac{32}{50}$

$\frac{16}{25}$

Hence we found that \[P\left( A|B \right)=\frac{16}{25}\]

3. If $P\left( A \right)=0.8$, $P\left( B \right)=0.5$ and $\mathbf{P}\left( B|A \right)=0.4$ find

(i) $P\left( A\bigcap B \right)$

Ans: It is given that $P\left( A \right)=0.8$, $P\left( B \right)=0.5$, and \[P\left( B|A \right)=0.4\]

Put all the data in the following formula

\[\text{ }P\left( B|A \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}\]

$\Rightarrow P\left( A\bigcap B \right)=P\left( A|B \right).P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.4\times 0.8$

Thus we found that $P\left( A\bigcap B \right)=0.32$

(iii) $\mathbf{P}\left( A|B \right)$

Ans: Given in the question $P\left( A \right)=0.8$, $P\left( B \right)=0.5$, and \[P\left( B|A \right)=0.4\]

We know that \[P\left( A|B \right)\]is the probability of occurrence of A when B has already happened

$\therefore P\left( A|B \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

Now put $P\left( A\bigcap B \right)=0.32$, $P\left( B \right)=0.5$in the above equation as shown

$\Rightarrow P\left( A|B \right)=\frac{0.32}{0.5}$

$=0.64$

Thus we found that $P\left( A|B \right)=0.64$

(iii) $\mathbf{P}\left( A\bigcup B \right)$

Ans: Given in the question $P\left( A \right)=0.8$, $P\left( B \right)=0.5$, $P\left( B \right)=0.5$and \[P\left( B|A \right)=0.4\]

Now we have the formula as shown

$\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

Put $\text{ }P\left( A \right)=0.8$,

$\text{ }P\left( B \right)=0.5$,

$\text{ }P\left( A\bigcap B \right)=0.32$ in the above as shown

$\therefore P\left( A\bigcup B \right)=0.8+0.5-0.32$

$\Rightarrow P\left( A\bigcup B \right)=0.98$

Thus we found that $P\left( A\bigcup B \right)=0.98$

4. Evaluate $\mathbf{P}\left( A\bigcup B \right)$ if $2\mathbf{P}\left( A \right)=P\left( B \right)=\frac{5}{13}$ and $P\left( A\left| B \right. \right)=\frac{2}{5}$

Ans: It is Given that

$2P\left( A \right)=P\left( B \right)=\frac{5}{13}$ and

$P\left( A\left| B \right. \right)=\frac{2}{5}$

$\Rightarrow P\left( A \right)=\frac{5}{26}$

Now we know the formula

$\text{ }P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

$\Rightarrow \text{ }\frac{2}{5}=\frac{13P\left( A\bigcap B \right)}{5}$ (since $P\left( B \right)=\frac{5}{13}$, $P\left( A\left| B \right. \right)=\frac{2}{5}$)

$\Rightarrow P\left( A\bigcap B \right)=\frac{2}{13}$

Also $P\left( A\bigcup B \right)$ is given by the formula

$\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow P\left( A\bigcup B \right)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$

$\Rightarrow P\left( A\bigcup B \right)=\frac{11}{26}$

Thus we found that $P\left( A\bigcup B \right)=\frac{11}{26}$

5. If $\mathbf{P}\left( A \right)=\frac{6}{11}$, $\mathbf{P}\left( B \right)=\frac{5}{11}$ and $\mathbf{P}\left( A\bigcup B \right)=\frac{7}{11}$ find

(i) $\mathbf{P}\left( A\bigcap B \right)$

Ans: Given $P\left( A \right)=\frac{6}{11}$, $P\left( B \right)=\frac{5}{11}$

Also it is given that

$P\left( A\bigcup B \right)=\frac{7}{11}$

And we know that it is given by the formula

$\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\text{ }\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P\left( A\bigcap B \right)$

$\Rightarrow P\left( A\bigcap B \right)=\frac{11}{11}-\frac{7}{11}$

$\Rightarrow P\left( A\bigcap B \right)=\frac{4}{11}$

Thus we found that $P\left( A\bigcap B \right)=\frac{4}{11}$

(ii) $\mathbf{P}\left( A\left| B \right. \right)$

Ans: Given $P\left( A \right)=\frac{6}{11}$, $P\left( B \right)=\frac{5}{11}$

Also we know that $P\left( A\left| B \right. \right)$ is given by

$\text{ }P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

\[\Rightarrow P\left( A\left| B \right. \right)=\frac{4}{11}\times \frac{11}{5}\] (since $P\left( A\bigcap B \right)=\frac{4}{11}$)

Thus we found that \[P\left( A\left| B \right. \right)=\frac{4}{5}\]

(iii) $\mathbf{P}\left( B\left| A \right. \right)$

Ans: Given $P\left( A \right)=\frac{6}{11}$, $P\left( B \right)=\frac{5}{11}$

Also we know that $P\left( B\left| A \right. \right)$ is given by

$\text{ }P\left( B\left| A \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}$

$\Rightarrow P\left( B\left| A \right. \right)=\frac{4}{11}\times \frac{11}{6}$

Thus we found that $P\left( B\left| A \right. \right)=\frac{2}{3}$

6. A coin is tossed three times, where

(i) E: head on third toss, F: heads on first two tosses

Ans: Sample space is given by

$S=\left\{ \left. HHH,HHT,HTH,HTT,THH,THT,TTH,TTT \right\} \right.$and the events E and F and their probabilities are given by

$E=\left\{ \left. HHH,HTH,THH,TTH \right\} \right.$

Therore $P\left( E \right)=\frac{4}{8}$

$F=\left\{ \left. HHH,HHT \right\} \right.$

Therore $P\left( F \right)=\frac{2}{8}$

$\therefore E\bigcap F=\left\{ \left. HHH \right\} \right.$

Therore $P\left( E\bigcap F \right)=\frac{1}{8}$

And hence $P\left( E\left| F \right. \right)$ is given by

$\text{ }P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

\[\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{1}{8}}{\frac{2}{8}}\]

Thus \[P\left( E\left| F \right. \right)=\frac{1}{2}\]

(ii) E: at least two heads, F: at most two heads

Ans: Sample space is given by

$S=\left\{ \left. HHH,HHT,HTH,HTT,THH,THT,TTH,TTT \right\} \right.$and the events E and F and their probabilities are given by

$E=\left\{ \left. HHH,HHT,THH,HTH \right\} \right.$

Therore $P\left( E \right)=\frac{4}{8}$

$F=\left\{ \left. HHT,HTH,HTT,THH,THT,TTH,TTT \right\} \right.$

Therore $P\left( F \right)=\frac{7}{8}$

$E\bigcap F=\left\{ \left. HHT,HTH,THH \right\} \right.$

Therore $P\left( E\bigcap F \right)=\frac{3}{8}$

And hence $P\left( E\left| F \right. \right)$ is given by

$\text{ }P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

\[\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{3}{8}}{\frac{7}{8}}\]

Thus \[P\left( E\left| F \right. \right)=\frac{3}{7}\]

(iii) E: at most two tails, F: at least one tail.

Ans: Sample space is given by

$S=\left\{ \left. HHH,HHT,HTH,HTT,THH,THT,TTH,TTT \right\} \right.$and the events E and F and their probabilities are given by

$E=\left\{ \left. HHH,HHT,THH,HTH,THH,THT,TTH \right\} \right.$

Therore $P\left( E \right)=\frac{7}{8}$

$F=\left\{ \left. HHT,HTT,HTH,THH,THT,TTH,TTT \right\} \right.$

Therore $P\left( F \right)=\frac{7}{8}$

$E\bigcap F=\left\{ \left. HHT,HTT,HTH,THH,THT,TTH \right\} \right.$

Therore $P\left( E\bigcap F \right)=\frac{6}{8}$

And hence $P\left( E\left| F \right. \right)$ is given by

$\text{ }P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

\[\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{6}{8}}{\frac{7}{8}}\]

Thus \[P\left( E\left| F \right. \right)=\frac{6}{7}\]

7. Two coins are tossed once, where

(i) E: tail appears on one coin, F: one coin shows head

Ans: Sample Space is given by

$S=\left\{ \left. HH,HT,TH,TT \right\} \right.$

The events E and F and their probabilities are given by

$E=\left\{ \left. HT,TH \right\} \right.$

Therefore $P\left( E \right)=\frac{2}{4}$

$F=\left\{ \left. HT,TH \right\} \right.$

Therefore $P\left( F \right)=\frac{2}{4}$

Also $E\bigcap F$ is given by

$E\bigcap F=\left\{ \left. HT,TH \right\} \right.$

We know that $P\left( E\left| F \right. \right)$ is given by

\[\text{ }P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}\]

$\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{2}{4}}{\frac{2}{4}}$

Thus we found that $P\left( E\left| F \right. \right)=1$

That is, $\left( E\left| F \right. \right)$ is a sure event

(ii) E: not tail appears, F: no head appears

Ans: Sample Space is given by

$S=\left\{ \left. HH,HT,TH,TT \right\} \right.$

The events E and F and their probabilities are given by

$E=\left\{ \left. HH \right\} \right.$

Therefore $P\left( E \right)=\frac{1}{4}$

$F=\left\{ \left. TT \right\} \right.$

Therefore $P\left( F \right)=\frac{1}{4}$

Therefore $E\bigcap F$ is given by

$E\bigcap F=\phi $

We know that $P\left( E\left| F \right. \right)$ is given by

\[\text{ }P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}\]

$\Rightarrow P\left( E\left| F \right. \right)=\frac{0}{\frac{1}{4}}$

Thus we found that $P\left( E\left| F \right. \right)=0$

8. A die is thrown three times,

E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses

Ans: Number of elements in Sample space is given by $216$

The events E and F and their probabilities are given by

$E=\left\{ \left. \begin{align} & \left( 1,1,4 \right),\left( 1,2,4 \right)......\left( 1,6,4 \right) \\ & \left( 2,1,4 \right),\left( 2,2,4 \right)........\left( 2,6,4 \right) \\ & \left( 3,1,4 \right),\left( 3,2,4 \right)..........\left( 3,6,4 \right) \\ & \left( 4,1,4 \right),\left( 4,2,4 \right)..........\left( 4,6,4 \right) \\ & \left( 5,1,4 \right),\left( 5,2,4 \right)...........\left( 5,6,4 \right) \\ & \left( 6,1,4 \right),\left( 6,2,4 \right)...........\left( 6,6,4 \right) \\ \end{align} \right\} \right.$

$F=\left\{ \left. \left( 6,5,1 \right),\left( 6,5,2 \right),\left( 6,5,3 \right),\left( 6,5,4 \right),\left( 6,5,5 \right),\left( 6,5,6 \right) \right\} \right.$

Therefore $P\left( F \right)=\frac{6}{216}$

And hence $E\bigcap F=\left\{ \left. \left( 6,5,4 \right) \right\} \right.$

Therefore $P\left( E\bigcap F \right)=\frac{1}{216}$

We know that $P\left( E\left| F \right. \right)$ is given by

$\text{ }P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

$\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{1}{216}}{\frac{6}{216}}$

Thus $P\left( E\left| F \right. \right)=\frac{1}{6}$

9. Mother, father and son line up at random for a family picture

E: son on one end, F: father in middle

Ans: Let mother (M), father (F), and son (S) line up for the family picture, then the sample space will be as shown

$A=\left( \left. MFS,MSF,FMS,FSM,SMF,SFM \right\} \right.$

The events E and F and their probabilities are

$E=\left\{ \left. MFS,FMS,SMF,SFM \right\} \right.$

$F=\left\{ \left. MFS,SFM \right\} \right.$

Therefore $P\left( F \right)=\frac{2}{6}$

Hence $E\bigcap F=\left\{ \left. MFS,SFM \right\} \right.$

Therefore $P\left( E\bigcap F \right)=\frac{2}{6}$

We know that $P\left( E\left| F \right. \right)$ is given by

$\Rightarrow P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

$\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{4}{6}}{\frac{4}{6}}$

Thus $P\left( E\left| F \right. \right)=1$

10. A black and a red dice are rolled.

Find the conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$.

Ans: Let the first observation come from black die and second from red die respectively

In the case when two dices are rolled the elements in sample space is $36$

The events A and B and their probabilities are given by

\[A=\left\{ \left( 4,6 \right),\left( 5,5 \right),\left( 5,6 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 4,6 \right)\left( 6,6 \right) \right\}\]

Where A is the event when sum is greater than $9$

Similarly,

\[B=\left\{ \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right) \right\}\]

Where B is the event when black die resulted in a $5$

Therefore $P\left( B \right)=\frac{6}{36}$

And hence $A\bigcap B=\left\{ \left( 5,5 \right),\left( 5,6 \right) \right\}$

Therefore the conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$$P\left( A\left| B \right. \right)$ is given by

$\Rightarrow P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

$\Rightarrow P\left( A\left| B \right. \right)=\frac{\frac{2}{36}}{\frac{6}{36}}$

Thus $P\left( A\left| B \right. \right)=\frac{1}{3}$

Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Ans: Let E and F be the events and their probabilities defined as

E: Sum of the observations is $8$

$E=\left\{ \left( 2,6 \right),\left( 3,5 \right),\left( 4,4 \right),\left( 5,3 \right),\left( 6,2 \right) \right\}$

F: red die resulted in a number less than $4$

\[F=\left\{ \begin{align} & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 2,1 \right),\left( 2,2 \right),\left( 2,2 \right) \\ & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right) \\ & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right) \\ \end{align} \right\}\]

Therefore $P\left( F \right)=\frac{18}{36}$

And hence $E\bigcap F=\left\{ \left( 5,3 \right),\left( 6,2 \right) \right\}$

Therefore $P\left( E\bigcap F \right)=\frac{2}{36}$

The conditional probability of obtaining the sum $8$, given that the red die resulted in a number less than $4$is given $P\left( E\left| F \right. \right)$ as shown

$P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{p\left( F \right)}$

$\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{2}{36}}{\frac{18}{36}}$

Thus $P\left( E\left| F \right. \right)=\frac{1}{9}$

11. A fair die is rolled. Consider events E = {1, 3, 5), F = {2, 3} and G = {2, 3, 4, 5}. Find

(i) $\mathbf{P}\left( E\left| F \right. \right)$ and $\mathbf{P}\left( F\left| E \right. \right)$

Ans: The sample space is given by

$S=\left\{ 1,2,3,4,5,6 \right\}$

It is given in the question

$E=\left\{ 1,3,5 \right\}$

$F=\left\{ 2,3 \right\}$

Therefore $E\bigcap F=\left\{ 3 \right\}$

Hence $P\left( E\bigcap F \right)=\frac{1}{6}$

$\therefore P\left( E \right)=\frac{3}{6}$

$\therefore P\left( F \right)=\frac{2}{6}$

Hence $P\left( E\left| F \right. \right)$ is given by

$P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

$\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{1}{6}}{\frac{2}{6}}$

Similarly $P\left( F\left| E \right. \right)$ si given by

$P\left( F\left| E \right. \right)=\frac{\frac{1}{6}}{\frac{3}{6}}$

Thus $P\left( E\left| F \right. \right)=\frac{1}{2}$ and $P\left( F\left| E \right. \right)=\frac{1}{3}$

(ii). $\mathbf{P}\left( E\left| G \right. \right)$ and $\mathbf{P}\left( G\left| E \right. \right)$

It is given in the question

$E=\left\{ 1,3,5 \right\}$

$\therefore P\left( E \right)=\frac{3}{6}$

$G=\left\{ 2,3,4,5 \right\}$

$\therefore P\left( G \right)=\frac{4}{6}$

$\therefore E\bigcap G=\left\{ 3,5 \right\}$

$\therefore P\left( E\bigcap G \right)=\frac{2}{6}$

Therefore $P\left( E\left| G \right. \right)=\frac{\left( E\bigcap G \right)}{P\left( G \right)}$

$\Rightarrow P\left( E\left| G \right. \right)=\frac{\frac{2}{6}}{\frac{4}{6}}$

Thus $P\left( E\left| G \right. \right)=\frac{1}{2}$

Similarly

$P\left( G\left| E \right. \right)=\frac{\frac{2}{6}}{\frac{3}{6}}$

Thus $P\left( G\left| E \right. \right)=\frac{2}{3}$

Therefore, $P\left( E\left| G \right. \right)=\frac{1}{2}$ and $P\left( G\left| E \right. \right)=\frac{2}{3}$

(iii) $\mathbf{P}\left( \left( E\bigcup G \right)\left| G \right. \right)$ and $\mathbf{P}\left( \left( E\bigcap G \right)\left| G \right. \right)$

Ans: The sample space is given by

$S=\left\{ 1,2,3,4,5,6 \right\}$

$G=\left\{ 2,3,4,5 \right\}$

We have $E\bigcup F=\left\{ 1,2,3,5 \right\}$

Therefore \[\begin{align} & \left( E\bigcup F \right)\bigcap G=\left\{ 1,2,3,5 \right\}\bigcap \left\{ 2,3,4,5 \right\} \\ & \text{ =}\left\{ 2,3,5 \right\} \\ \end{align}\]

Also \[\begin{align} & \left( E\bigcap F \right)\bigcap G=\left\{ 1,2,3,5 \right\}\bigcap \left\{ 3 \right\} \\ & \text{ =}\left\{ 3 \right\}\text{ } \\ \end{align}\]

$\therefore P\left( E\bigcup F \right)\bigcap G=\frac{3}{6}$

$\therefore P\left( E\bigcap F \right)\bigcap G=\frac{1}{6}$

$\therefore P\left( E\bigcup F\left| G \right. \right)=\frac{P\left( E\bigcup F \right)\bigcap G}{P\left( G \right)}$

$\Rightarrow P\left( E\bigcup F\left| G \right. \right)=\frac{\frac{3}{6}}{\frac{4}{6}}$

Thus $P\left( E\bigcup F\left| G \right. \right)=\frac{3}{4}$

Similarly

$P\left( E\bigcap F\left| G \right. \right)=\frac{\frac{1}{6}}{\frac{4}{6}}$

Thus, $P\left( E\bigcup F\left| G \right. \right)=\frac{3}{4}$ and $P\left( E\bigcap F\left| G \right. \right)=\frac{1}{4}$

12. Assume that each born child is equally like to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

i. The youngest is a girl

Ans: The sample space for a family having two children is given by

$S=\left\{ \left( B,B \right),\left( B,G \right),\left( G,G \right),\left( G,B \right) \right\}$

Where B refers to boy child and G refers to girl child

Let an event be defined as

E: Both children are girls

$E=\left\{ \left( GG \right) \right\}$

$\therefore P\left( E \right)=\frac{2}{4}$

Let F be an event defined as

F: youngest child is girl

$F=\left\{ \left( BG \right),\left( GG \right) \right\}$

$\therefore E\bigcap F=\left\{ \left( GG \right) \right\}$

$\therefore P\left( E\bigcap F \right)=\frac{1}{4}$

We know that $P\left( E\left| F \right. \right)$ is given by

$P\left( E\left| F \right. \right)=\frac{P\left( E\bigcap F \right)}{P\left( F \right)}$

$\Rightarrow P\left( E\left| F \right. \right)=\frac{\frac{1}{4}}{\frac{2}{4}}$

Thus $P\left( E\left| F \right. \right)=\frac{1}{2}$

ii. At least one is a girl

Ans: The sample space for a family having two children is given by

$S=\left\{ \left( B,B \right),\left( B,G \right),\left( G,G \right),\left( G,B \right) \right\}$

Where B refers to boy child and G refers to girl child

Let an event be defined as

E: Both children are girls

$E=\left\{ \left( GG \right) \right\}$

$\therefore P\left( E \right)=\frac{2}{4}$

Let A be an event defined as

$\text{ }A=\left\{ \left( B,G \right),\left( G,B \right),\left( G,G \right) \right\}$

$\therefore P\left( A \right)=\frac{3}{4}$

$\therefore E\bigcap A=\left\{ \left( G,G \right) \right\}$

$\therefore P\left( E\bigcap A \right)=\frac{1}{4}$

We know that

$P\left( E\left| A \right. \right)$ is given by

$P\left( E\left| A \right. \right)=\frac{P\left( \bigcap A \right)}{P\left( A \right)}$

$\Rightarrow P\left( E\left| A \right. \right)=\frac{\frac{1}{4}}{\frac{3}{4}}$

Thus $P\left( E\left| A \right. \right)=\frac{1}{3}$

13. An instructor has a bank consisting of 300 easy True/False s, 200 difficult True/False s, 500 easy multiple choice s and 400 difficult multiple choice s. If a is selected at random from the bank, what is the probability that it will be an easy given that it is a multiple choice ?

Ans: The given data can be represented as

True/False | Multiple choice | Total | |

Easy | 300 | 500 | 800 |

Difficult | 200 | 400 | 600 |

Total | 500 | 900 | 1400 |

Let us have following notations

E for easy questions, M for multiple questions, D for difficult questions and T for true/false questions

It is given that total number of questions is $1400$, that of multiple questions is $900$

Hence probability for selecting easy multiple choice questions is given by

$P\left( E\bigcap M \right)=\frac{5}{14}$ (since $\frac{500}{1400}=\frac{5}{14}$)

Probability for selecting multiple choice questions is given by

$P\left( M \right)=\frac{9}{14}$ (since $\frac{90}{1400}=\frac{9}{14}$)

$\therefore P\left( E\left| M \right. \right)=\frac{P\left( \bigcap M \right)}{P\left( M \right)}$

$\Rightarrow P\left( E\left| M \right. \right)=\frac{\frac{5}{14}}{\frac{9}{14}}$

Thus $P\left( E\left| M \right. \right)=\frac{5}{9}$

14. Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is $4$’.

Ans: Let A and be events defined as

A:the sum of the numbers on the dice is $4$

B:the two numbers appearing on throwing the two dice are different.

$\therefore A=\left\{ \left( 1,3 \right),\left( 2,3 \right),\left( 3,1 \right) \right\}$

$\therefore B=\left\{ \begin{align} & \left( 1,2 \right),\left( 1,3 \right)....\left( 1,6 \right) \\ & \left( 2,1 \right),\left( 2,2 \right)....\left( 2,6 \right) \\ & \left( 3,1 \right),\left( 3,2 \right)......\left( 3,6 \right) \\ & \left( 4,1 \right),\left( 4,2 \right).......\left( 4,6 \right) \\ & \left( 5,1 \right),\left( 5,2 \right).........\left( 5,6 \right) \\ & \left( 6,1 \right),\left( 6,2 \right).........\left( 6,6 \right) \\ \end{align} \right\}$

Therefore $P\left( B \right)=\frac{30}{36}$

$\therefore A\bigcap B=\left\{ \left( 1,3 \right),\left( 3,1 \right) \right\}$

$P\left( A\bigcap B \right)=\frac{2}{36}$

We know that $P\left( A\left| B \right. \right)$ is given by

$P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

$\Rightarrow P\left( A\left| B \right. \right)=\frac{\frac{2}{36}}{\frac{30}{36}}$

Thus \[P\left( A\left| B \right. \right)=\frac{1}{15}\]

15. Consider the experiment of throwing a die, if a multiple of $3$ comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows as $3$’.

Ans: For this case the sample space is given by

$S=\left\{ \begin{align} & \left( 1,H \right),\left( 2,H \right),\left( 1,T \right),\left( 1,T \right),\left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right) \\ & \left( 4,H \right),\left( 4,T \right),\left( 5,H \right),\left( 5,T \right),\left( 6,1 \right),\left( 6,2 \right)......\left( 6,6 \right) \\ \end{align} \right\}$

Let A and B be events defined as

A:The coin shows tail

B: at least one die shows $3$

$\therefore A=\left\{ \left( 1,T \right),\left( 2,T \right),\left( 4,T \right),\left( 5,T \right) \right\}$

$\therefore B=\left\{ \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right),\left( 6,3 \right) \right\}$

$\therefore P\left( B \right)=\frac{7}{36}$

Also it is observable that $A\bigcap B=\phi $

We know that $P\left( A\left| B \right. \right)$ given by

$P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

$P\left( A\left| B \right. \right)=\frac{0}{\frac{7}{36}}$

Thus $P\left( A\left| B \right. \right)=0$

16. If $P\left( A \right)=\frac{1}{2}$and $P\left( B \right)=0$then $P\left( A\left| B \right. \right)$is

$0$

$\frac{1}{2}$

Not defined

$1$

Ans: It is given that $P\left( A \right)=\frac{1}{2}$ and $P\left( B \right)=0$

We know that $P\left( A\left| B \right. \right)$ is given by

$P\left( A\left| B \right. \right)=\frac{P\left( \bigcap B \right)}{P\left( B \right)}$

$\Rightarrow P\left( A\left| B \right. \right)=\frac{P\left( \bigcap B \right)}{0}$

Thus $P\left( A\left| B \right. \right)$ is not defined

17. If A and B are events such that $P\left( A\left| B \right. \right)=P\left( B\left| A \right. \right)$ then

(A)$A\subset but\text{ A}\ne \text{B}$ (B) $A=B$ (C)$A\bigcap B=\phi $) (D)= $P\left( A \right)=P\left( B \right)$

Ans: It is given in the question that

$\text{ }P\left( A\left| B \right. \right)=P\left( B\left| A \right. \right)$

$\Rightarrow \frac{P\left( A\bigcap B \right)}{P\left( B \right)}=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}$

$\text{ }\Rightarrow \frac{P\left( A \right)}{1}=\frac{P\left( B \right)}{1}$

Thus we found that $P\left( A \right)=P\left( B \right)$

### Exercise 13.2

1. if $P\left( A \right)=\frac{3}{5}$ and $P\left( B \right)=\frac{1}{5}$, find $P\left( A\bigcap B \right)$ if A and B are independent events

Ans: It is given in the question that $P\left( A \right)=\frac{3}{5}$ and $P\left( B \right)=\frac{1}{5}$

Also it is given that A and B are independent events

$\text{ }\therefore P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=\frac{3}{5}\times \frac{1}{5}$

Thus $P\left( A\bigcap B \right)=\frac{3}{25}$

2. Two cards are drawn at random and without replacement from a pack of $2$ playing cards. Find the probability that both the cards are black.

Ans: Given a deck of $52$ cards

We know that the number of black cards are $26$

Let us have a notation $P\left( A \right)$ denote the probability of picking a black card at first chance

$\therefore P\left( A \right)=\frac{26}{52}$

Similarly, let us have a notation $P\left( B \right)$ denote the probability of picking a black card at second chance.

Since there is no replacement

$\therefore P\left( B \right)=\frac{25}{51}$

The probability that the both the cards are black is

$P\left( A\bigcap B \right)=\frac{1}{2}\times \frac{25}{51}$

$=\frac{25}{102}$

The probability that the both the cards are black is $\frac{25}{102}$

3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.

Ans: Let A,B and C be the events and their probabilities defined as

A: the first orange is good

$\therefore P\left( A \right)=\frac{12}{15}$

B: the second orange is good

$\therefore P\left( B \right)=\frac{1}{15}$

C: the third orange is good

$\therefore P\left( C \right)=\frac{10}{15}$

It is given that the box is approved for sale only when all the oranges are good.

Probability that the box is approved for sale=probability of all oranges to be good

Therefore probability of all oranges to be good=$\frac{12}{15}\times \frac{11}{15}\times \frac{10}{15}$

Thus probability that the box is approved for sale=$\frac{44}{91}$

4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event $3$ on the die’. Check whether A and B are independent events or not.

Ans: The sample space is given by

$S=\left\{ \begin{align} & \left( H,1 \right),\left( H,2 \right)......\left( H,6 \right) \\ & \left( T,1 \right),\left( T,2 \right)........\left( T,6 \right) \\ \end{align} \right\}$

According to the question

$A=\left\{ \left( H,1 \right),\left( H,2 \right)......\left( H,6 \right) \right\}$

$\therefore P\left( A \right)=\frac{6}{12}$

\[B=\left\{ \left( H,3 \right),\left( T,3 \right) \right\}\]

$\therefore P\left( B \right)=\frac{2}{12}$

Also $\left( A\bigcap B \right)=\left\{ \left( H,3 \right) \right\}$

$\therefore P\left( A\bigcap B \right)=\frac{1}{12}$

We know that for A and B to be independent events

$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now $P\left( A \right)P\left( B \right)=\frac{6}{12}\times \frac{2}{12}$

$\Rightarrow P\left( A \right)P\left( B \right)=\frac{1}{12}$

Which is as same as $P\left( A\bigcap B \right)$

Thus A and B are independent events

5. A die marked $1,2,3$ in red and $4,5,6$in green is tossed. Let A be the events, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

Ans: The sample space is given by

$S=\left\{ 1,2,3,4,5,6 \right\}$

According to the question

$A=\left\{ 2,4,6 \right\}$

$\therefore P\left( A \right)=\frac{3}{6}$

\[B=\left\{ 1,2,3 \right\}\]

$\therefore P\left( B \right)=\frac{3}{6}$

Also $\left( A\bigcap B \right)=\left\{ 2 \right\}$

$\therefore P\left( A\bigcap B \right)=\frac{1}{6}$

We know that for A and B to be independent events

$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now $P\left( A \right)P\left( B \right)=\frac{3}{6}\times \frac{3}{6}$

$\Rightarrow P\left( A \right)P\left( B \right)=\frac{1}{4}$

Which is not as same as $P\left( A\bigcap B \right)$

Thus A and B are not independent events

6. Let E and F be the vents with $P\left( E \right)=\frac{3}{5}$, $P\left( F \right)=\frac{3}{10}$ and $P\left( E\bigcap F \right)=\frac{1}{5}$

Are E and F independent?

Ans: Given $P\left( E \right)=\frac{3}{5}$, $P\left( F \right)=\frac{3}{10}$ and $P\left( E\bigcap F \right)=\frac{1}{5}$

We know that for A and B to be independent events

$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now $P\left( A \right)P\left( B \right)=\frac{3}{5}\times \frac{3}{10}$

$\Rightarrow P\left( A \right)P\left( B \right)=\frac{9}{50}$

Also $P\left( E\bigcap F \right)=\frac{1}{5}$

Thus $P\left( A \right)P\left( B \right)\ne P\left( E\bigcap F \right)$

And hence A and B are not independent events

7. Given that the events A and B are such that $P\left( A \right)=\frac{1}{2}$, $P\left( A\bigcup B \right)=\frac{3}{5}$ and $P\left( B \right)=p$ find p if they are

(i) Mutually Exclusive

Ans: It is given in the question that $P\left( A \right)=\frac{1}{2}$, $P\left( B \right)=p$ and $P\left( A\bigcup B \right)=\frac{3}{5}$

We know that if two events A and B are mutually exclusive then $P\left( A\bigcap B \right)=0$

Also we know that

$P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0$

$\Rightarrow p=\frac{3}{5}-\frac{1}{2}$

$\Rightarrow p=\frac{1}{10}$

(ii) Independent

Ans: It is given in the question that $P\left( A \right)=\frac{1}{2}$, $P\left( B \right)=p$ and $P\left( A\bigcup B \right)=\frac{3}{5}$

We know that if two events A and B are independent then $P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Also we know that

$P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}$ (since $P\left( A \right)P\left( B \right)=\frac{p}{2}$)

$\Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}$

$\Rightarrow p=\frac{2}{10}$

$=\frac{1}{5}$

The value of $p$ when A and B are independent event is $\frac{1}{5}$

8. Let A and B be independent events with $P\left( A \right)=0.3$ and $P\left( B \right)=0.4$ find

(i) $P\left( A\bigcap B \right)$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then

$\text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

(ii) $P\left( A\bigcup B \right)$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then

$\text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Also we know that

$P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow P\left( A\bigcup B \right)=0.3+0.4-0.12$

$\Rightarrow P\left( A\bigcup B \right)=0.58$

(iii) $P\left( A\left| B \right. \right)$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then

$\text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Also we know that $P\left( A\left| B \right. \right)$ is given by

$P\left( A\left| B \right. \right)=\frac{\left( P\left( A\bigcap B \right) \right)}{P\left( B \right)}$

$\Rightarrow P\left( A\left| B \right. \right)=\frac{0.12}{0.4}$

$=0.3$

Thus we found that $P\left( A\left| B \right. \right)=0.3$

(iv) $P\left( B\left| A \right. \right)$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then

$\text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Also we know that \[P\left( B\left| A \right. \right)\] is given by

$P\left( B\left| A \right. \right)=\frac{\left( P\left( A\bigcap B \right) \right)}{P\left( A \right)}$

$\Rightarrow P\left( B\left| A \right. \right)=\frac{0.12}{0.3}$

$=0.4$

Thus we found that $P\left( B\left| A \right. \right)=0.4$

9. If A and B are two events such that $P\left( A \right)=\frac{1}{4}$, $P\left( B \right)=\frac{1}{2}$ and $P\left( A\bigcap B \right)=\frac{1}{8}$ find $P\left( \text{not A and not B} \right)$

Ans: It is given that two events A and b have probabilities as shown

$P\left( A \right)=\frac{1}{4}$, $P\left( B \right)=\frac{1}{2}$ and \[P\left( A\bigcap B \right)=\frac{1}{8}\]

$P\left( \text{not A and not B} \right)=P\left( A'\bigcap B' \right)$

And we know that

$\text{ }P\left( A'\bigcap B' \right)=P\left( A\bigcup B \right)'$

$\text{ }P\left( A'\bigcap B' \right)=1-P\left( A\bigcup B \right)$

Also $\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\text{ }\Rightarrow P\left( A\bigcup B \right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$

\[\text{ }\Rightarrow P\left( A\bigcup B \right)=\frac{5}{8}\]

$\therefore P\left( A'\bigcap B' \right)=1-\frac{5}{8}$

Thus $P\left( \text{not A and not B} \right)=\frac{3}{8}$

10. Events A and B are such that $P\left( A \right)=\frac{1}{2}$, $P\left( B \right)=\frac{7}{12}$ and $P\left( \text{not A or not B} \right)=\frac{1}{4}$. State whether A and B are independent

Ans: It is given that $P\left( A \right)=\frac{1}{2}$, \[P\left( B \right)=\frac{7}{12}\] and \[P\left( A'\bigcup B' \right)=\frac{1}{4}\]

According to the question

$1-P\left( A\bigcap B \right)=\frac{1}{4}$

$\Rightarrow P\left( A\bigcap B \right)=\frac{3}{4}$

We know that A and B are independent events if

$P\left( A \right)P\left( B \right)=P\left( A\bigcap B \right)$

$\therefore P\left( A \right)P\left( B \right)=\frac{1}{2}\times \frac{7}{12}$

Clearly $P\left( A \right)P\left( B \right)\ne P\left( A\bigcap B \right)$

Thus A and B are not independent events

11. Given two independent events A and B such that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$. Find

(i) $P\left( A\text{ and B} \right)$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

$\therefore \text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( \text{A and B} \right)=0.3\times 0.6$

Thus we found that $P\left( \text{A and B} \right)=0.18$

(ii) $P\left( A\text{ and not B} \right)$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

$\therefore P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now we know that

$P\left( \text{A and not B} \right)=P\left( A\bigcap B' \right)$

Also we know that

$\text{ }P\left( A\bigcap B' \right)=P\left( A \right)-P\left( A\bigcap B \right)$

$\text{ }P\left( A\bigcap B' \right)=0.3-0.18$

Thus we found that $P\left( \text{A and not B} \right)=0.12$

(iii) $P\left( A\text{ or B} \right)$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

$\therefore \text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( \text{A and B} \right)=0.3\times 0.6$

Thus we found that $P\left( \text{A and B} \right)=0.18$

Also we know that

$\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\text{ }P\left( A\bigcup B \right)=0.3+0.6-0.18$

Thus we know that $P\left( A\text{ or B} \right)=0.72$

(iv) $P\left( neither\text{ }A\text{ and nor B} \right)$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

$\therefore \text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( \text{A and B} \right)=0.3\times 0.6$

Thus we found that $P\left( \text{A and B} \right)=0.18$

Also we know that

$\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\text{ }P\left( A\bigcup B \right)=0.3+0.6-0.18$

Thus we know that $P\left( A\text{ or B} \right)=0.72$

$P\left( \text{neither A nor B} \right)=P\left( A\bigcup B \right)'$

And $\text{ }P\left( A\bigcup B \right)'=1-P\left( A\bigcup B \right)$

$\text{ }P\left( A\bigcup B \right)'=1-0.72$

Thus $P\left( \text{neither A nor B} \right)=0.28$

12. A die tossed thrice. Find the probability of getting an odd number at least once.

Ans: It is given that a die is tossed thrice, the number of elements in sample space for each throw is $6$

Now probability of getting atleast odd once=1-probability of getting even numbers in all three events

Now we know that

Probability of getting even numbers in all three events=${{\left( \frac{1}{2} \right)}^{3}}\text{ }$

Probability of getting at least odd once=1-$\frac{1}{8}$

Thus probability of getting at least odd once=$\frac{7}{8}$

13. Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls.

Find the probability that

(i) Both balls are red.

Ans: It is given in the question that

Number of red balls =8 and number of black balls =10

Now probability of getting red ball at first draw,say P(A)=$\frac{8}{18}$=$\frac{4}{9}$

Also probability of getting red ball at second draw, say (P)=$\frac{8}{18}$=$\frac{4}{9}$

$\therefore P\left( A\bigcap B \right)=\frac{4}{9}\times \frac{4}{9}$(since the events are independent)

$=\frac{16}{81}$

The probability of getting both balls red is $\frac{16}{81}$

(ii) First ball is black and the second is red.

Ans: It is given in the question that

Number of red balls =8 and number of black balls =10

Now probability of getting black ball at first draw=$\frac{10}{18}$=$\frac{5}{9}$

Also probability of getting red ball at second draw=$\frac{8}{18}$=$\frac{4}{9}$

Also above two events are independent therefore their intersection is given by

$=\frac{5}{9}\times \frac{4}{9}$

$=\frac{20}{81}$

$\begin{align} & \text{probability of getting black ball in first draw} \\ & \text{and red ball in second draw =}\frac{20}{81} \\ \end{align}$

(iii) One of them is black and the other is red.

Ans: Let us have following notations

A: first ball is black and the other is red

Now probability of getting a black ball in first draw, $P\left( A \right)=\frac{10}{18}$

$=\frac{5}{9}$

As the ball is replaced after first throw, probability of getting a red ball in second draw $=\frac{8}{18}$

$=\frac{4}{9}$

Probability of getting first ball as black and second ball as red $=\frac{5}{9}\times \frac{4}{9}$

$=\frac{20}{81}$

Probability of drawing red ball in first draw $=\frac{8}{18}$

$=\frac{4}{9}$

Probability of getting black ball in second draw$=\frac{10}{81}$

$=\frac{5}{9}$

Probability of drawing first ball as red and second as black $=\frac{5}{9}\times \frac{4}{9}$

$=\frac{20}{81}$

Probability of getting one of them is black and the other is red$=\frac{20}{81}+\frac{20}{81}$

$=\frac{40}{81}$

Probability of getting one of the ball is black and the other is red is $\frac{40}{81}$

14. The probability of solving specific problems independently by A and B are $\frac{1}{2}$and $\frac{1}{3}$respectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved

Ans: Given that the two events A and B are independent with probabilities given as shown

$P\left( A \right)=\frac{1}{2}$ and $P\left( B \right)=\frac{1}{3}$

$\therefore \text{ }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=\frac{1}{6}$

Probability of problem being solved is given by

$\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

\[\Rightarrow P\left( A\bigcup B \right)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\]

Thus probability of problem being solved is $\frac{2}{3}$

(ii) exactly one of them solves the problem

Ans: Given that the two events A and B are independent with probabilities given as shown

$P\left( A \right)=\frac{1}{2}$ and $P\left( B \right)=\frac{1}{3}$

Either one of them solved the answer. So either A solve the answer but not B or B solved the answer but not A

$P\left( A \right).P\left( B' \right)+P\left( A' \right).P\left( B \right)$

$=\frac{1}{2}\left( \frac{2}{3} \right)+\frac{1}{2}\left( \frac{1}{3} \right)$

$=\frac{2}{6}+\frac{1}{6}$

$=\frac{3}{6}$

$=\frac{1}{2}$

Probability of one of them solves the answer is $\frac{1}{2}$

15. One card is drawn at random from a well shuffled deck of $52$ cards. In which of the following cases are the events E and F independent?

(i) E: ‘the card drawn is a spade’

F: ‘the card drawn is an ace’

Ans: Given a deck of $52$ cards

We know that $13$cards are spades and $4$are aces

$P\left( E \right)=\frac{13}{52}$ and $P\left( F \right)=\frac{4}{52}$

Also in the deck only $1$ card is ace of spade

I.e $P\left( E\bigcap F \right)=\frac{1}{52}$

It is clearly visible that

$P\left( E \right)P\left( F \right)=\frac{13}{52}\times \frac{4}{52}$ equals $P\left( E\bigcap F \right)$

Hence the events are independent

(ii)E: ‘the card drawn is black’

F: ‘the card drawn is a king’

Ans: We know that in the deck of cards $26$ cards are black and $4$ cards are kings

$P\left( E \right)=\frac{26}{52}$ and $P\left( F \right)=\frac{4}{52}$

Also there are 2 cards which are black as well as king

i.e $P\left( E\bigcap F \right)=\frac{2}{52}$

It is clearly visible that $\frac{26}{52}\times \frac{4}{52}=\frac{2}{52}$

Hence these are independent events

(iii) E: ‘the card drawn is a king and queen’

F: ‘the card drawn is a queen or jack’

Ans: In a deck of $52$ cards each king, queen and jack is $4$

$\therefore P\left( E \right)=P\left( F \right)=\frac{8}{52}$

There are $4$cards which which are king and queen or jack

Hence its probability is $\frac{4}{52}$

Clearly E and F are not independent

16. In a hostel, $60$ percent of the students read Hindi newspapers, $40$ percent read English newspapers and $20$ percent read both Hindi and English newspapers. A student is selected at random.

Find the probability that she reads neither Hindi nor English newspapers.

Ans: Let us have following notations

H: Students who read Hindi newspapers

E: Students who read English newspapers

$\therefore P\left( H \right)=0.6$ and

$\therefore P\left( E \right)=0.4$

Also it is given that

$P\left( H\bigcap E \right)=0.2$

Therefore the required probability is given by

\[P\left( H\bigcup E' \right)=1-P\left( H\bigcup E \right)\]

\[P\left( H\bigcup E' \right)=1-\left( P\left( H \right)+P\left( E \right)-P\left( H\bigcap E \right) \right)\]

\[P\left( H\bigcup E' \right)=1-0.8\]

Hence the probability that she reads neither Hindi nor English is $0.2$

If she reads Hindi news paper, find the probability that she reads English news paper.

Ans: Let us have following notations

H: Students who read Hindi newspapers

E: Students who read English newspapers

$\therefore P\left( H \right)=0.6$ and

$\therefore P\left( E \right)=0.4$

Also it is given that

$P\left( H\bigcap E \right)=0.2$

Now required probability is given by

\[\text{ }P\left( E\left| H \right. \right)=\frac{P\left( H\bigcap E \right)}{P\left( H \right)}\]

\[\Rightarrow P\left( E\left| H \right. \right)=\frac{0.2}{0.6}\]

Thus the probability that she reads English newspapers if she reads Hindi news papers is $\frac{1}{3}$

If she reads English newspapers, find the probability that she reads Hindi newspapers.

Ans: Let us have following notations

H: Students who read Hindi newspapers

E: Students who read English newspapers

$\therefore P\left( H \right)=0.6$ and

$\therefore P\left( E \right)=0.4$

Also it is given that

$P\left( H\bigcap E \right)=0.2$

The required probability is given by

$\text{ }P\left( E\left| H \right. \right)=\frac{P\left( H\bigcap E \right)}{P\left( H \right)}$

$\Rightarrow P\left( E\left| H \right. \right)=\frac{0.2}{0.4}$

Therefore the probability that she reads Hindi newspapers provided she reads English newspapers is $\frac{1}{2}$

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

$0$

$\frac{1}{3}$

$\frac{1}{12}$

$\frac{1}{36}$

Ans: Out of six numbers in a die, the only even prime number is $2$

Therefore probability of getting an even prime number is given by

Say $P\left( E \right)=\frac{1}{36}$

Hence probability of getting an even prime number is $\frac{1}{36}$

18. Two events A and B will be independently, if

A and B are mutually exclusive

Ans: Let $P\left( A \right)=p$ and $P\left( B \right)=q$ $0<p,q<1$

It is given that they are mutually exclusive

$\therefore P\left( A\bigcap B \right)=0$

But $P\left( A \right)P\left( B \right)=pq$ which is not necessarily zero

Hence they are not independent

\[P\left( A'B' \right)=\left[ 1-P\left( A \right) \right]\left[ 1-P\left( B \right) \right]\]

Ans: Two events are said to be independent if $P\left( AB \right)=P\left( A \right)P\left( B \right)$

Let us solve for option B first

$P\left( A'B' \right)=\left[ 1-P\left( A \right) \right]\left[ 1-P\left( B \right) \right]$

$\Rightarrow P\left( A'\bigcap B' \right)=1-P\left( A \right)-P\left( B \right)+P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)=P\left( A \right)+P\left( B \right)-P\left( A \right)P\left( B \right)$

Hence we found that $P\left( AB \right)=P\left( A \right)P\left( B \right)$

Hence these are independent events

\[P\left( A \right)=P\left( B \right)\]

Ans: If $P\left( A \right)=P\left( B \right)$ then

Each of them has probability of $\frac{1}{2}$

Surely it can be sometime true but not a sufficient condition for two events to be independent

\[P\left( A \right)+P\left( B \right)=1\]

Ans: Similarly If $P\left( A \right)+P\left( B \right)=1$ it is not necessary condition for two events to be independent

### Exercise 13.3

1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Ans: It is given in the question that at random one ball is picked and its colour is noted then two balls of same colour is added, hence we have two cases here

Consider the first case

Let the first ball drawn be Red, its probability is $\frac{5}{10}$

Now two extra red balls are added,

$\therefore$ P (drawing red ball at second attempt =$\frac{7}{12}$

Since these two events are independent therefore required probability for this case is $\frac{5}{10}\times \frac{7}{12}$

Consider the second case

Let black ball be drawn at first attempt, its probability is $\frac{5}{10}$

Now two extra black balls will be added

$\therefore$ P(drawing red ball at second attempt)=$\frac{5}{12}$

Since these two events are independent therefore required probability for this case is $\frac{5}{10}\times \frac{5}{12}$

Total probability is given by $\frac{5}{10}\times \frac{7}{12}+\frac{5}{10}\times \frac{5}{12}=\frac{5}{10}$

Thus the probability of drawing second ball red is $\frac{1}{2}$

2. A bag contains 4 red and 4 black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Ans: Let E and F be the events of selecting first and second bag respectively

$\therefore P\left( E \right)=P\left( F \right)=\frac{1}{2}$

Let K be the event of getting a red ball

$\therefore P\left( K\left| E \right. \right)=\frac{4}{8}$, where

$\left( K\left| E \right. \right)$: drawing red ball from the first bag

Similarly, $P\left( K\left| F \right. \right)=\frac{2}{8}$, where

$\left( K\left| F \right. \right)$: drawing red ball from the first bag

Now from the Bay’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}\]

Thus the probability that the red ball is drawn from the first bag is$\frac{2}{3}$

3. Of the students in a college, it is known that $60$ percent reside in a hostel and $40$ percent are day scholars (not residing in a hostel). Previous year results report that $30$ percent of all students who reside in hostels attain A grade and $20$ percent of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is hostile?

Ans: Let us have following notations

E: The event when a student is a hostler

F: The event when a student is a day scholar

K: The event chosen student gets grade A

$\therefore P\left( E \right)=0.6$

$\therefore P\left( F \right)=0.4$

$\therefore P\left( K\left| E \right. \right)=0.3$

$\therefore P\left( K\left| F \right. \right)=0.2$

By Bay’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}\]

Thus the probability that the gets grade A provided he Is a hostler is $\frac{9}{13}$

4. In answering a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?

Ans: Let us have following notations

E: The event when a student knows the answer

F: The event when a student guesses the answer

K: The event that the answer is correct

$\therefore P\left( E \right)=\frac{3}{4}$

$\therefore P\left( F \right)=\frac{1}{4}$

$\therefore P\left( K\left| E \right. \right)=1$ (since if he knows the answer then he will answer it correctly, it’s a trivial case)

where $\left( K\left| E \right. \right)$: the answer is correct provided he knows the answer

$\therefore P\left( K\left| F \right. \right)=\frac{1}{4}$

$\left( K\left| F \right. \right)$: the answer is correct provided he guesses the answer

By Bay’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{3}{4}\times \frac{1}{1}}{\frac{3}{4}\times \frac{1}{1}+\frac{1}{4}\times \frac{1}{4}}\]

Thus the probability that he knows the answer provided he answered correctly is $\frac{12}{13}$

5. A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present.

However, the test also yields a false positive result for $0.5$ percent of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If $0.1$ percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Ans: Let us have following notations

E: The event when the person has disease

F: The event when the person does not have disease

K: The event that the test result is positive

$\therefore P\left( E \right)=0.001$

It is clear that E and F are complementary events

$\therefore P\left( E \right)+P\left( F \right)=1$

Hence

$P\left( F \right)=0.999$

$\therefore P\left( K\left| E \right. \right)=0.99$

$\left( K\left| E \right. \right)$: the result is correct provided he has the disease

$\therefore P\left( K\left| F \right. \right)=0.005$

$\left( K\left| F \right. \right)$: the result is correct provided he doesn’t have the disease

By Bay’s theorem

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}\]

Thus the probability that the person had the disease provided his test result is positive is $\frac{22}{133}$

6. There are three coins. One is a two headed coin (having heads on both faces), another is a biased coin that comes up heads $75$ percent of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Ans: Let us have following notations

E: The event when chosen coin is two-headed

F: The event when chosen coin is biased

G: The event when chosen coin is unbiased

K: The event that the coin shows heads

$\therefore P\left( E \right)=\frac{1}{3}$

$P\left( F \right)=\frac{1}{3}$

$P\left( G \right)=\frac{1}{3}$

$\therefore P\left( K\left| E \right. \right)=1$ (since it is a trivial case)

$\left( K\left| E \right. \right)$: the coins shows heads provided it is two headed

$\therefore P\left( K\left| F \right. \right)=\frac{3}{4}$

$\left( K\left| F \right. \right)$: the coins shows heads provided it is biased

$\therefore P\left( K\left| G \right. \right)=\frac{1}{2}$

$\left( K\left| G \right. \right)$: the coins shows heads provided it is unbiased

By Baye’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)+P\left( G \right)P\left( K\left| G \right. \right)}$

\[P\left( E\left| K \right. \right)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}\]

Thus the probability that the coin shows heads provided it is two headed coin is $\frac{4}{9}$

7. An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of accidents are $0.01$ , $0.03$ and $0.15$respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Ans: Let us have following notations

E: The event when the driver drives scooter

F: The event when the driver drives car

G: The event when the driver drives truck

K: The event that the driver meets accident

$\therefore P\left( E \right)=\frac{2000}{12000}$

$P\left( F \right)=\frac{4000}{12000}$

$P\left( G \right)=\frac{6000}{12000}$

$\therefore P\left( K\left| E \right. \right)=\frac{1}{100}$

$\left( K\left| E \right. \right)$: the driver meets accident provided he drives scooter

$\therefore P\left( K\left| F \right. \right)=\frac{3}{100}$

$\left( K\left| F \right. \right)$: the driver meets accident provided he drives car

$\therefore P\left( K\left| G \right. \right)=\frac{15}{100}$

$\left( K\left| G \right. \right)$: the driver meets accident provided he drives truck

By Baye’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)+P\left( G \right)P\left( K\left| G \right. \right)}$

\[P\left( E\left| K \right. \right)=\frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{1}{2}\times \frac{15}{100}}\]

Thus the probability that the accidented person is a scooter driver is $\frac{1}{52}$

8.A factory has two machines A and B. Past record shows that machine A produced $60$ percent of the items of output and machine B produced $40$ percent of the items. Future, $2$ percent of the items produced by machine A and $1$ percent produced by machine B were defective. All the items are put into one stockpile and then one item is chosen randomly from this and is found to be defective. What is the probability that it was produced by machine B?

Ans: Let us have following notations

E: The event when the production is from machine A

F: The event when the production is from machine B

K: The event that the item is defective

$\therefore P\left( E \right)=0.6$

$\therefore P\left( F \right)=0.4$

$\therefore P\left( K\left| E \right. \right)=0.02$

$\left( K\left| E \right. \right)$: The event that the item is defective provided it from A

$\therefore P\left( K\left| F \right. \right)=0.01$

$\left( K\left| F \right. \right)$: The event that the item is defective provided it from B

By Baye’s theorem

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{0.4\times 0.01}{0.6\times 0.02+0.4\times 0.01}\]

Thus the probability that that the item is defective provided it from B is $\frac{1}{4}$

9. Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are $0.6$and $0.4$respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$and the corresponding probability is $0.3$if the second group wins. Find the probability that the new product introduced was by the second group.

Ans: Let us have following notations

E: The event when the first group wins

F: The event when the second group wins

K: The event that the new item is produced

$\therefore P\left( E \right)=0.6$

$\therefore P\left( F \right)=0.4$

$\therefore P\left( K\left| E \right. \right)=0.7$

$\left( K\left| E \right. \right)$: The event that the item is introduced by first group

$\therefore P\left( K\left| F \right. \right)=0.3$

$\left( K\left| F \right. \right)$: The event that item is introduced by second group

By Baye’s theorem

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3}\]

Thus the probability that that the item is produced from second group is $\frac{2}{9}$

10. Suppose a girl throws a die. If she gets $5$ or $6$, she tosses a coin three times and notes the number of heads. If she gets $1,2,3,\text{ or 4}$, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw $1,2,3,\text{ or 4}$ with the die?

Ans: Let us have following notations

E: The event when $5\text{ or 6}$ come

F: The event when $\text{1,2,3 or 4}$ come

K: The event that the coin shows exactly one head

$\therefore P\left( E \right)=\frac{2}{6}$

$\therefore P\left( F \right)=\frac{4}{6}$

$\therefore P\left( K\left| E \right. \right)=\frac{3}{8}$

$\left( K\left| E \right. \right)$: The event that the coin shows exactly one head provided $5\text{ or 6}$come

$\therefore P\left( K\left| F \right. \right)=\frac{1}{2}$

$\left( K\left| F \right. \right)$: The event that the single throw of coin shows exactly one head provided $\text{1,2,3 or 4}$come

By Baye’s theorem

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{\frac{4}{6}\times \frac{1}{2}}{\frac{2}{6}\times \frac{3}{8}+\frac{4}{6}\times \frac{1}{2}}\]

Thus the probability that that the item is produced from second group is $\frac{8}{11}$

11. A manufacturer has three machine operators A, B and C. The first operator A produces$1$ percent defective items, whereas the other two operators B and C produce $5$ percent and $7$ percent defective items respectively. A is on the job for $50$ Percent of the time, B is on the job for $30$ Percent of the time and C is on the job for $20$ percent of the time. A defective item is produced, what is the probability that it was produced by A?

Ans: Let us have following notations

E: The event when A works

F: The event when B works

G: The event when C works

K: The event that the defective item is produced

$\therefore P\left( E \right)=0.5$

$\therefore P\left( F \right)=0.3$

$\therefore P\left( G \right)=0.2$

$\therefore P\left( K\left| E \right. \right)=0.01$

$\left( K\left| E \right. \right)$: The event that the defective item is produced from A

$\therefore P\left( K\left| F \right. \right)=0.05$

$\left( K\left| F \right. \right)$: The event that the defective item is produced from B

$\therefore P\left( K\left| G \right. \right)=0.07$

The event that the defective item is produced from C

By Baye’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)+P\left( G \right)\times P\left( K\left| G \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{0.5\times 0.01}{0.5\times 0.01+0.3\times 0.05+0.2\times 0.07}\]

Thus the probability that that the item is produced from A and found defective $\frac{5}{34}$

12. A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Ans: Let us have following notations

E: The event when a card is diamond

F: The event when the card is not diamond

K: The event to denote lost card

Clearly E and F are contemporary Events

$\therefore P\left( E \right)=\frac{13}{52}$ (since $13$cards are diamonds)

$\therefore P\left( F \right)=\frac{39}{52}$

$\left( K\left| E \right. \right)$: The event of drawing two cards when one diamond card is lost

Since one diamond card is lost therefore left total cards are

$51$ and left diamond cards are $13$

$\therefore P\left( K\left| E \right. \right)=\frac{^{12}{{C}_{2}}}{^{51}{{C}_{2}}}$

$\Rightarrow P\left( K\left| E \right. \right)=\frac{22}{425}$

$\left( K\left| F \right. \right)$: The event of drawing two cards when one non diamond card is lost

Since one card is lost therefore left total cards are $51$

$\therefore P\left( K\left| F \right. \right)=\frac{^{13}{{C}_{2}}}{^{51}{{C}_{2}}}$

$\Rightarrow P\left( K\left| F \right. \right)=\frac{26}{425}$

By Baye’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{1}{4}\times \frac{22}{425}}{\frac{1}{4}\times \frac{22}{425}+\frac{3}{4}\times \frac{26}{425}}\]

Thus the probability of the lost card being a diamond is $\frac{11}{50}$

13. Probability that A speaks the truth is $\frac{4}{5}$. A coin is tossed. A report that a head appears. The probability that actually there was head is

$\frac{4}{5}$

$\frac{1}{2}$

$\frac{1}{5}$

$\frac{2}{5}$

Ans: Let us have following notations

E: The event when A speaks truth

F: The event when A speak false

K: The event it is head

Clearly E and F are contemporary Events

$\therefore P\left( E \right)=\frac{4}{5}$ (since $13$cards are diamonds)

$\therefore P\left( F \right)=\frac{1}{5}$

$\left( K\left| E \right. \right)$: The event it is head and he speaks truth

$\therefore P\left( K\left| E \right. \right)=\frac{1}{2}$

$\left( K\left| F \right. \right)$: The event it is head and he speaks false

$\therefore P\left( K\left| F \right. \right)=\frac{1}{2}$

By Baye’s theorem

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{4}{5}\times \frac{1}{2}}{\frac{4}{5}\times \frac{1}{2}+\frac{1}{5}\times \frac{1}{2}}\]

Thus the probability of the lost card being a diamond is $\frac{4}{5}$

14. If A and B are two events such that $A\subset B$ and $P\left( B \right)\ne 0$then which of the following is correct

$P\left( A\left| B \right. \right)=\frac{P\left( B \right)}{P\left( A \right)}$

$P\left( A\left| B \right. \right)<P\left( A \right)$

$P\left( A\left| B \right. \right)\ge P\left( A \right)$

None of these

Ans: Given $A\subset B$

$\text{ }\therefore A\bigcap B=A$ and $P\left( A \right)<P\left( B \right)$

Now $P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

$\Rightarrow P\left( A\left| B \right. \right)=\frac{P\left( A \right)}{P\left( B \right)}$

Now $P\left( B \right)\le 1$

$\therefore \frac{1}{P\left( B \right)}\ge 1$

Hence option C is correct, i.e $P\left( A\left| B \right. \right)\ge P\left( A \right)$

### Exercise 13.4

1. State which of the following are not the probability distribution of a random variable. Give reasons for your answer.

X | $0$ | $1$ | $2$ |

P(X) | $0.4$ | $0.4$ | $0.2$ |

Ans: for any distribution to be probability distribution $\sum{P\left( X \right)}=1$and $0\le P\left( X \right)\le 1$

$\because 0.4+0.4+0.2=1$

Therefore it is a probability distribution

X | $0$ | $1$ | $2$ | $3$ | $4$ |

P(X) | $0.1$ | $0.5$ | $0.2$ | $-0.1$ | $0.3$ |

Ans: for any distribution to be probability distribution $\sum{P\left( X \right)}=1$ and $0\le P\left( X \right)\le 1$

$\because -0.1$does not lie in the range

Therefore it is not a probability distribution

Y | $-1$ | $0$ | $1$ |

P(Y) | $0.6$ | $0.1$ | $0.2$ |

Ans: For any distribution to be probability distribution $\sum{P\left( X \right)}=1$and $0\le P\left( X \right)\le 1$

$\because 0.6+0.1+0.2\ne 1$

Therefore it is not a probability distribution

Z | $3$ | $2$ | $1$ | $0$ | $-1$ |

P(Z) | $0.3$ | $0.2$ | $0.4$ | $0.1$ | $0.05$ |

Ans: For any distribution to be probability distribution $\sum{P\left( X \right)}=1$and $0\le P\left( X \right)\le 1$

$\because 0.3+0.2+0.1+0.05\ne 1$

Therefore it is not a probability distribution

2. An urn contains $5$ red and $2$ black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

Ans: Let us have following notations

B: represents black ball

R: represents red ball

The sample space can be represented as

$S=\left\{ \left( BB \right),\left( BR \right),\left( RB \right),\left( RR \right) \right\}$

X represent the number of black ball

Thus possible values for X is $0,1,2$

Also $\sum{P\left( X \right)}=1$ it can be clearly seen

3. Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed $6$ times. What are possible values of X ?

Ans: Given that a coin is tossed six times and X denotes the difference between number of heads and that of tails

Mathematically we can write as shown

$\begin{align} & X\left( 6H,0T \right)=6 \\ & X\left( 5H,1T \right)=4 \\ & X\left( 4H,2T \right)=2 \\ & X\left( 3H,3T \right)=0 \\ & X\left( 2H,4T \right)=2 \\ & X\left( 1H,5T \right)=4 \\ & X\left( 0H,6T \right)=6 \\ \end{align}$

The possible values of X are $0,2,4,6$

4. Find the probability distribution of

(i) number of heads in two tosses of a coin

Ans: the sample space is given by

$S=\left\{ \left( HH \right),\left( HT \right),\left( TH \right),\left( TT \right) \right\}$

Let Y represents the number of heads then

$Y\left( HH \right)=2$, $Y\left( HT \right)=1$, $Y\left( TH \right)=1$, $Y\left( TT \right)=0$

Also since the coins are unbiased therefore each of them has probability of $\frac{1}{4}$

We have

$P\left( Y=0 \right)=\frac{1}{4}$

$P\left( Y=1 \right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$P\left( Y=2 \right)=\frac{1}{4}$

Therefore table made is as shown

Y | $0$ | $1$ | $2$ |

P(Y) | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ |

(ii) number of tails in the simultaneous tosses of three coins

Ans: the sample space is given by

$S=\left\{ \left( HHH \right),\left( HTH \right),\left( HTT \right),\left( THH \right),\left( THT \right),\left( TTH \right),\left( TTT \right) \right\}$

Let Y represents the number of tails then it can take values $0,1,2,\text{ }or\text{ }3$

Also since the coins are unbiased therefore each of them has probability of $\frac{1}{4}$

We have

$P\left( Y=0 \right)=\frac{1}{8}$

$P\left( Y=1 \right)=\frac{3}{8}$

$P\left( Y=2 \right)=\frac{3}{8}$

$P\left( Y=3 \right)=\frac{1}{8}$

Therefore table made is as shown

(Y) | $0$ | $1$ | $2$ | $3$ |

P(Y) | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |

(iii) number of heads in four tosses of a coin

Ans: The sample space is given by

$S=\left\{ \begin{align} & \left( HHHH \right),\left( HHHT \right),\left( HHTT \right),\left( HTHT \right),\left( HTHH \right),\left( HTTH \right),\left( HTTT \right) \\ & \left( THHH \right),\left( THHT \right),\left( THTH \right),\left( THTT \right),\left( TTHH \right),\left( TTHT \right),\left( TTTH \right),\left( TTTT \right) \\ \end{align} \right\}$

Let Y represents the number of heads then it can take values $0,1,2,3,\text{ }or\text{ }4$

Also since the coins are unbiased therefore each of them has probability of $\frac{1}{4}$

We have

$P\left( Y=0 \right)=\frac{1}{16}$

$P\left( Y=1 \right)=\frac{4}{16}$

\[P\left( Y=2 \right)=\frac{6}{16}\]

$P\left( Y=3 \right)=\frac{4}{16}$

$P\left( Y=4 \right)=\frac{1}{16}$

Therefore table made is as shown

Y | $0$ | $1$ | $2$ | $3$ | $4$ |

P(Y) | $\frac{1}{16}$ | $\frac{4}{16}$ | \[\frac{6}{16}\] | $\frac{4}{16}$ | $\frac{1}{16}$ |

5. Find the probability distribution of the number of success in two tosses of die, where a success is defined as

(i) number greater than 4

Ans: Let Y be the random variable which represents the number of successes i.e it represents number greater than $4$

We can easily observe that Y can have three values as shown

$0$ for number less than or equal to $4$in the both tosses

$1$ for greater than $4$ and less than or equal to $4$ in either of tosses

$2$ for greater than $4$in the both tosses

Now $P\left( Y=0 \right)$ means probability of number less than $4$

$\therefore P\left( Y=0 \right)=\frac{4}{6}\times \frac{4}{6}=\frac{4}{9}$

$\therefore P\left( Y=1 \right)=\frac{4}{6}\times \frac{2}{6}+\frac{2}{6}\times \frac{4}{6}=\frac{4}{9}$

$\therefore P\left( Y=2 \right)=\frac{2}{6}\times \frac{2}{6}=\frac{1}{9}$

Y | $0$ | $1$ | $2$ |

P(Y) | $\frac{4}{9}$ | $\frac{4}{9}$ | $\frac{1}{9}$ |

(ii) Six appears on at least one die

Ans: Let Y be the random variable which represents the number of successes i.e it represents six appearing at least on one die

We can easily observe that Y can have two values as shown

$0$ for when six doesn’t appear in any of the dices

$1$ for when six appears in either of the dice

Now $P\left( Y=0 \right)$ means probability six doesn’t appear in any of the dices

$\therefore P\left( Y=0 \right)=\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$

$\therefore P\left( Y=1 \right)=\frac{1}{6}\times \frac{5}{6}+\frac{5}{6}\times \frac{1}{6}=\frac{5}{18}$

Y | $0$ | $1$ |

P(Y) | $\frac{25}{36}$ | $\frac{5}{18}$ |

6. From a list of $30$ bulbs which includes $6$ defectives, a sample of $4$ bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Ans: Let us denote D to be the random variable that denotes the number of defective bulbs

Number of defective bulbs is $30-6=24$

Let us have following notations

$D=0$ means all four selected are non defective

$D=1$ means three selected are non defective and one is defective

$D=2$ means two selected are non defective and two is defective

$D=3$ means one selected are non defective and three is defective

$D=4$ means all four selected are defective

\[\therefore P\left( D=0 \right){{=}^{4}}{{C}_{0}}{{\left( \frac{4}{5} \right)}^{4}}=\frac{256}{625}\]

\[\therefore P\left( D=1 \right){{=}^{4}}{{C}_{1}}{{\left( \frac{4}{5} \right)}^{3}}\frac{1}{5}=\frac{256}{625}\]

\[\therefore P\left( D=2 \right){{=}^{4}}{{C}_{2}}{{\left( \frac{4}{5} \right)}^{2}}{{\left( \frac{1}{5} \right)}^{2}}=\frac{96}{625}\]

\[\therefore P\left( D=3 \right){{=}^{4}}{{C}_{1}}\left( \frac{4}{5} \right){{\left( \frac{1}{5} \right)}^{3}}=\frac{16}{625}\]

\[\therefore P\left( D=4 \right){{=}^{4}}{{C}_{4}}{{\left( \frac{1}{5} \right)}^{4}}=\frac{1}{625}\]

D | $0$ | $1$ | $2$ | $3$ | $4$ |

P(D) | \[\frac{256}{625}\] | \[\frac{256}{625}\] | \[\frac{96}{625}\] | \[\frac{16}{625}\] | \[\frac{1}{625}\] |

7. A coin is biased so that the head is $3$ times as likely to occur as the tail. If the coin is tossed twice, find the probability distribution of the number of tails.

Ans: Given that the ratio of occurring heads is $3$times that of tails

i.e $P\left( H \right)=3P\left( T \right)$

Also we know that

$\Rightarrow P\left( T \right)=\frac{1}{4}$

$\Rightarrow P\left( H \right)=\frac{3}{4}$

The sample space for toss of two coins is given by

$S=\left\{ \left( HH \right),\left( HT \right),\left( TH \right),\left( TT \right) \right\}$

Let K be the random variable which represents number of tails

Therefore possible values of K are $0,1,2$

$\therefore P\left( K=0 \right)=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

$\therefore P\left( K=1 \right)=\frac{3}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{3}{4}=\frac{6}{16}$

$\therefore P\left( K=2 \right)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

Hence the distributions is

K | $0$ | $1$ | $2$ |

P(K) | $\frac{9}{16}$ | $\frac{6}{16}$ | $\frac{1}{16}$ |

8. A random variable X has the following probability distribution.

Y | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |

P(Y) | $0$ | $k$ | $2k$ | $2k$ | $3k$ | ${{k}^{2}}$ | $2{{k}^{2}}$ | $7{{k}^{2}}+k$ |

Determine

(i)k

Ans: We know that the sum of probabilities is $1$

$\begin{align} & \Rightarrow k+2k+2k+3k+{{k}^{2}} \\ & +2{{k}^{2}}+7{{k}^{2}}+k=1 \\ \end{align}$

$\Rightarrow 10{{k}^{2}}+9k-1=0$

$\left( 10k-1 \right)\left( k+1 \right)=0$

$\Rightarrow k=-1,\frac{1}{10}$

(ii) $P\left( X<3 \right)$

Ans: we know that

$P\left( X<3 \right)=P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right)$

$P\left( X<3 \right)=0+k+2k=3k$

$P\left( X<3 \right)=\frac{3}{10}$

(iii) $P\left( X>6 \right)$

Ans: Since there are $7$possible values for random variable

$\therefore P\left( X>6 \right)=P\left( X=7 \right)$

$\Rightarrow P\left( X=7 \right)=7{{k}^{2}}+k$

$\Rightarrow P\left( X=7 \right)=7{{\left( \frac{1}{10} \right)}^{2}}+\frac{1}{10}$

$\Rightarrow P\left( X>6 \right)=\frac{17}{100}$

(iv) $P\left( 0<X<3 \right)$

Ans: we know that

$\text{ }P\left( 0<X<3 \right)=P\left( X=1 \right)+P\left( X=2 \right)$

$\Rightarrow P\left( 0<X<3 \right)=k+2k$

$\Rightarrow P\left( 0<X<3 \right)=\frac{3}{10}$

9. The random variable X has probability P(X) of the following form, where k is some number:

$P\left( X \right)=\left\{ \begin{align} & k,\text{ }if\text{ }x=0 \\ & 2k,\text{ }if\text{ }x=1 \\ & 3k,\text{ }if\text{ }x=2 \\ & 0,\text{ otherwise} \\ \end{align} \right\}$

Determine the value of k.

Ans: We know that total sum of probabilities equal to $1$

i.e $k+2k+3k=1$

$\text{ }\Rightarrow k=\frac{1}{6}$

Find $P\left( X<2 \right)$,$P\left( X\le 2 \right)$ , $P\left( X\ge 2 \right)$.

Ans: We know that

$P\left( X<2 \right)=P\left( X=0 \right)+P\left( X=1 \right)$

$\Rightarrow P\left( X<2 \right)=\frac{3}{6}$

Similarly,

$P\left( X\le 2 \right)=P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right)$

$\Rightarrow P\left( X\le 2 \right)=\frac{6}{6}$

Similarly,

$P\left( X\ge 2 \right)=P\left( X=2 \right)+0$

$\Rightarrow P\left( X\ge 2 \right)=\frac{3}{6}$

10: Find the mean number of heads in three tosses of a fair coin.

Ans: The sample is given by

$S=\left\{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT \right\}$

Let K denotes getting heads

It is clearly visible that K can take values $0,1,2,3$

$P\left( K=0 \right)=P\left( TTT \right)$

$P\left( K=0 \right)=P\left( T \right)P\left( T \right)P\left( T \right)$

$P\left( K=0 \right)=\frac{1}{8}$

Similarly,

$P\left( K=1 \right)=\frac{3}{8}$

$P\left( K=2 \right)=\frac{3}{8}$

$P\left( K=3 \right)=\frac{1}{8}$

The probability distribution is

(K) | $0$ | $1$ | $2$ | $3$ |

P(K) | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |

Mean is given by

$E\left( K \right)=\sum{{{K}_{i}}P\left( {{K}_{i}} \right)}$

$\Rightarrow E\left( K \right)=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$

Thus mean value is $1.5$

11. Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Ans: Here, we take X to represent the number of sixes obtained when two dice are thrown simultaneously.

Hence, it is easily visible that possible values for X \[0,1,2\]

\[0\] represents number of zero sixes in any of the dices

\[1\] represents number of sixes in any one of the dices

\[2\] represents number of zero sixes in both of the dices

$\therefore P\left( X=0 \right)={{\left( \frac{5}{6} \right)}^{2}}=\frac{25}{36}$

$\therefore P\left( X=1 \right)=2\left( \frac{5}{6} \right)\left( \frac{1}{6} \right)=\frac{10}{36}$

$\therefore P\left( X=2 \right)={{\left( \frac{1}{6} \right)}^{2}}=\frac{1}{36}$

Thus, the required probability distribution is as follows.

X | $0$ | $1$ | $2$ |

P(X) | $\frac{25}{36}$ | $\frac{10}{36}$ | $\frac{1}{36}$ |

Mean or Expectation is given by

$E\left( K \right)=\sum{{{K}_{i}}P\left( {{K}_{i}} \right)}$

$\Rightarrow E\left( K \right)=0\times \frac{25}{36}+1\times \frac{10}{36}+2\times \frac{1}{36}$

Thus mean value is $\frac{1}{3}$

12. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of two numbers obtained. Find E(X).

Ans: The ways in which two positive integers can be selected from first six without replacement is $6\times 5=30$ways

Let K denote the larger of two numbers obtained

Clearly it is visible that K can take values $2,3,4,5,6$

For $K=2$ possible observations are $\left( 1,2 \right)\text{ and }\left( 2,1 \right)$

$\therefore P\left( K=2 \right)=\frac{2}{30}$

For $K=3$ possible observations are $\left( 1,3 \right),\left( 3,1 \right),\left( 3,2 \right),\left( 2,3 \right)$

$\therefore P\left( K=3 \right)=\frac{4}{30}$

For $K=4$ possible observations are $\left( 1,4 \right),\left( 4,1 \right),\left( 4,2 \right),\left( 2,4 \right),\left( 3,4 \right),\left( 4,3 \right)$

$\therefore P\left( K=4 \right)=\frac{6}{30}$

For $K=5$ possible observations are $\left( 1,5 \right),\left( 5,1 \right),\left( 5,2 \right),\left( 2,5 \right),\left( 3,5 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 4,5 \right)$

$\therefore P\left( K=5 \right)=\frac{8}{30}$

For $K=6$ possible observations are $\left( 1,6 \right),\left( 6,1 \right),\left( 6,2 \right),\left( 2,6 \right),\left( 3,6 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 4,6 \right),\left( 5,6 \right),\left( 6,5 \right)$

$\therefore P\left( K=5 \right)=\frac{10}{30}$

Thus the required probability distribution is

K | $2$ | $3$ | $4$ | $5$ | $6$ |

P(K) | $\frac{2}{30}$ | $\frac{4}{30}$ | $\frac{6}{30}$ | $\frac{8}{30}$ | $\frac{10}{30}$ |

Hence $E\left( K \right)=\sum{{{K}_{i}}P\left( {{K}_{i}} \right)}$

$\Rightarrow E\left( K \right)=2\times \frac{2}{30}+3\times \frac{4}{30}+4\times \frac{6}{30}+5\times \frac{8}{30}+6\times \frac{10}{30}$

Thus expected value is $\frac{14}{3}$

13. Let X denote the sum of the number obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Ans: Let K be the random variable that denote the sum of numbers obtained when two dice rolled simultaneously

Clearly K can take values $2,3,4,5,6,7,8,9,10,11,12$

We have

$P\left( X=2 \right)=\frac{1}{36}$

$P\left( X=3 \right)=\frac{2}{36}$

$P\left( X=4 \right)=\frac{3}{36}$

$P\left( X=5 \right)=\frac{4}{36}$

$P\left( X=6 \right)=\frac{5}{36}$

$P\left( X=7 \right)=\frac{6}{36}$

$P\left( X=8 \right)=\frac{5}{36}$

$P\left( X=9 \right)=\frac{4}{36}$

$P\left( X=10 \right)=\frac{3}{36}$

$P\left( X=11 \right)=\frac{2}{36}$

$P\left( X=12 \right)=\frac{1}{36}$

And hence the probability distribution is

K | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ |

P(K) | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ |

Then $E\left( K \right)=\sum{{{K}_{i}}P\left( {{K}_{i}} \right)}$

$\begin{align} & \Rightarrow E\left( K \right)=2\times \frac{1}{36}+3\times \frac{2}{36}+4\times \frac{3}{36}+5\times \frac{4}{36}+6\times \frac{5}{36} \\ & +7\times \frac{6}{36}+8\times \frac{5}{36}+9\times \frac{4}{36}+10\times \frac{3}{36}+11\times \frac{2}{36}+12\times \frac{1}{36} \\ \end{align}$

$\Rightarrow E\left( K \right)=7$

Also $\begin{align} & E\left( {{K}^{2}} \right)=2\times {{\left( \frac{1}{36} \right)}^{2}}+3\times {{\left( \frac{2}{36} \right)}^{2}}+4\times {{\left( \frac{3}{36} \right)}^{2}}+5\times {{\left( \frac{4}{36} \right)}^{2}}+6\times {{\left( \frac{5}{36} \right)}^{2}} \\ & +7\times {{\left( \frac{6}{36} \right)}^{2}}+8\times {{\left( \frac{5}{36} \right)}^{2}}+9\times {{\left( \frac{4}{36} \right)}^{2}}+10\times {{\left( \frac{3}{36} \right)}^{2}}+11\times {{\left( \frac{2}{36} \right)}^{2}}+12\times {{\left( \frac{1}{36} \right)}^{2}} \\ \end{align}$

Hence $E\left( {{K}^{2}} \right)=\frac{329}{6}$

Now we know that variance is given by

\[E\left( {{K}^{2}} \right)-{{\left( E\left( K \right) \right)}^{2}}\]

$\begin{align} & =54.833-49 \\ & =5.833 \\ \end{align}$

And standard deviation is given by

$\sqrt{\operatorname{var}\left( K \right)}=\sqrt{5.833}$

Hence we found variance to be $5.833$ and standard deviation to be $2.415$

14. A class has $15$ students whose ages are $14,17,15,14,21,17,19,16,18,2017,16,19\text{ and 20}$ years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Ans: It is given that probability of choosing each is same i.e $\frac{1}{15}$

From the above data we can draw a frequency table as shown

X | $14$ | $15$ | $16$ | $17$ | $18$ | $19$ | $20$ | $21$ |

f | $2$ | $1$ | $2$ | $3$ | $1$ | $2$ | $3$ | $1$ |

We have following probabilities as shown

Hence probability distributions is as shown

X | $14$ | $15$ | $16$ | $17$ | $18$ | $19$ | $20$ | $21$ |

P(X) | $\frac{2}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{3}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{3}{15}$ | $\frac{1}{15}$ |

The mean is given by

$E\left( K \right)=\sum{{{K}_{i}}P\left( {{K}_{i}} \right)}$

$\begin{align} & \Rightarrow E\left( K \right)=14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+17\times \frac{3}{15} \\ & +18\times \frac{1}{15}+19\times \frac{2}{15}+20\times \frac{3}{15}+21\times \frac{1}{15} \\ \end{align}$

$\Rightarrow E\left( K \right)=17.53$

Now \[\begin{align} & E\left( {{K}^{2}} \right)={{\left( 14 \right)}^{2}}\times \frac{2}{15}+{{\left( 15 \right)}^{2}}\times \frac{1}{15}+{{\left( 16 \right)}^{2}}\times \frac{2}{15}+{{\left( 17 \right)}^{2}}\times \frac{3}{15} \\ & +{{\left( 18 \right)}^{2}}\times \frac{1}{15}+{{\left( 19 \right)}^{2}}\times \frac{2}{15}+{{\left( 20 \right)}^{2}}\times \frac{3}{15}+{{\left( 21 \right)}^{2}}\times \frac{1}{15} \\ \end{align}\]

$\Rightarrow E\left( {{K}^{2}} \right)=312.2$

Hence variance is given by

$E\left( {{K}^{2}} \right)-{{\left( E\left( K \right) \right)}^{2}}$

$=312.2-307.4177$

$=4.78$

Hence standard deviation is $\sqrt{4.78}=2.19$

15. In a meeting, $70$ percent of the members favour and $30$ percent oppose a certain proposal. A member is selected at random and we take $X=0$if he is opposed, and $X=1$if he is in favour. Find E(X) and var(X).

Ans: Given in the question that $X=0$means opposes the proposal and $X=1$ means that he is in favour.

$\therefore P\left( X=0 \right)=0.3$ and

$\text{ }P\left( X=1 \right)=0.7$

Thus the probability distribution is as shown

X | $0$ | $1$ |

P(X) | $0.3$ | $0.7$ |

We know that The mean is given by

$E\left( X \right)=\sum{{{X}_{i}}P\left( {{X}_{i}} \right)}$

$\Rightarrow E\left( X \right)=0\times 0.3+1\times 0.7$

$\Rightarrow E\left( X \right)=0.7$

Also $E\left( {{X}^{2}} \right)=0.7$

Now variance is given by

$E\left( {{X}^{2}} \right)-E\left( X \right)$

$=0.7-0.49$

$=0.21$

16. The mean of the numbers obtained on throwing a die having written $1$ on three faces, $2$ on two faces and $5$ on one face is

Ans: Let us denote K to be the random variable representing number of faces in which certain fixed number is to occur

According to the question

$\therefore P\left( K=1 \right)=\frac{3}{6}$

$\text{ }P\left( K=2 \right)=\frac{2}{6}$

$\text{ }P\left( K=5 \right)=\frac{1}{6}$

Thus the probability distribution is given by

K | $1$ | $2$ | $5$ |

P(K) | $\frac{3}{6}$ | $\frac{2}{6}$ | $\frac{1}{6}$ |

Thus mean is given by

$E\left( K \right)=\sum{{{K}_{i}}P\left( K \right)}$

$\Rightarrow E\left( K \right)=1\times \frac{3}{6}+2\times \frac{2}{6}+5\times \frac{1}{6}$

$\Rightarrow E\left( K \right)=2$

Hence we found that mean is $2$

17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is

Ans: Let us denote K to be the random variable which measures the number of aces occurred.

The possible values for K are $0,1,2$

$K=0$ means no ace is present

$K=1$ means one ace is present

$K=2$ means two aces are present

$\therefore P\left( K=0 \right)=\frac{^{4}{{C}_{0}}{{\times }^{48}}{{C}_{2}}}{^{52}{{C}_{2}}}=\frac{1128}{1326}$ (since there are $4$aces and $48$ non aces cards)

$\therefore P\left( K=1 \right)=\frac{^{4}{{C}_{1}}{{\times }^{48}}{{C}_{1}}}{^{52}{{C}_{2}}}=\frac{192}{1326}$

$\therefore P\left( K=2 \right)=\frac{^{4}{{C}_{2}}{{\times }^{48}}{{C}_{0}}}{^{52}{{C}_{2}}}=\frac{6}{1326}$

Hence the probability distribution is given by

K | $0$ | $1$ | $2$ |

P(K) | $\frac{1128}{1326}$ | $\frac{192}{1326}$ | $\frac{6}{1326}$ |

Thus mean is given by

$E\left( K \right)=\sum{{{K}_{i}}P\left( K \right)}$

$\Rightarrow E\left( K \right)=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}$

$\Rightarrow E\left( K \right)=\frac{2}{13}$

Hence we found that mean is $\frac{2}{13}$

### Exercise 13.5

1. A die is thrown $6$ times. If ‘getting an odd number’ is a success, what is the probability of

(i) $5$ successes

Ans: Let us have a notation K to denote to get success of getting odd numbers, number of trials being $6$

Probability of getting an odd number in a single throw is $p=\frac{3}{6}$

Probability of getting no odd number in a single throw is $q=1-\frac{3}{6}=\frac{3}{6}$

Hence we found $p=q=\frac{1}{2}$

Probability of getting $5$ successes is given by

$\text{ }P\left( K=5 \right){{=}^{6}}{{C}_{5}}{{\left( \frac{1}{2} \right)}^{6}}$

$\Rightarrow P\left( K=5 \right)=\frac{2\times 3}{{{2}^{6}}}$

$\Rightarrow P\left( K=5 \right)=\frac{3}{32}$

(ii) at least $5$ success

Ans: Let us have a notation K to denote to get success of getting odd numbers, number of trials being $6$

Probability of getting an odd number in a single throw is $p=\frac{3}{6}$

Probability of getting no odd number in a single throw is $q=1-\frac{3}{6}=\frac{3}{6}$

Hence we found $p=q=\frac{1}{2}$

Probability of getting at least $5$ successes is given by

\[\text{ }\sum\limits_{k=5}^{k=6}{P\left( K=k \right)}=\left( ^{6}{{C}_{5}}{{+}^{6}}{{C}_{6}} \right){{\left( \frac{1}{2} \right)}^{6}}\]

$\Rightarrow P\left( K=5 \right)+P\left( K=6 \right)=\frac{7}{64}$

(iii) at most $5$ successes?

Ans: Let us have a notation K to denote to get success of getting odd numbers, number of trials being $6$

Probability of getting an odd number in a single throw is $p=\frac{3}{6}$

Probability of getting no odd number in a single throw is $q=1-\frac{3}{6}=\frac{3}{6}$

Hence we found $p=q=\frac{1}{2}$

Probability of getting at most $5$ successes is given by

$1-P\left( K=6 \right)$

$=1-\frac{1}{64}$

Hence probability of getting at most $5$tails is $\frac{63}{64}$

2. A pair of dice is thrown $4$ times. If getting a doublet is considered a success, find the probability of two successes.

Ans: Let us have a notation K to be the random variable to denote getting a doublet as the success.

Probability of getting a doublet in a single throw $\text{ }p=\frac{6}{36}$

Probability of getting no doublet in a single throw $q=1-\frac{6}{36}=\frac{30}{36}$

Clearly K follows Binomial Distribution with

$n=4,p=\frac{6}{36},q=\frac{30}{36}$

\[\therefore P\left( K=2 \right){{=}^{4}}{{C}_{2}}\frac{{{5}^{4-2}}}{{{6}^{4}}}\]

\[\therefore P\left( K=2 \right)=\frac{25}{216}\]

Hence the probability of two success i.e getting $2$ doublets is \[\frac{25}{216}\]

3. There are $5$ percent defective items in a large bulk of items. What is the probability that a sample of $10$ items will include not more than one defective item?

Ans: Let us have a notation K to be the random variable which measures the number of defective items.

Given $\text{ }p=0.05$

Therefore $q=1-0.05=0.95$

Clearly K follows Binomial Distribution with

$n=10,p=\frac{1}{20},q=\frac{19}{20}$

Probability of getting not more than $1$ defective item is given by

$\text{ }P\left( K\le 1 \right)=P\left( K=0 \right)+P\left( K=1 \right)$

$\Rightarrow P\left( K\le 1 \right){{=}^{10}}{{C}_{0}}{{\left( \frac{1}{20} \right)}^{0}}{{\left( \frac{19}{20} \right)}^{10}}{{+}^{10}}{{C}_{1}}{{\left( \frac{1}{20} \right)}^{1}}{{\left( \frac{19}{20} \right)}^{9}}$

Hence we found that Probability of getting not more than $1$ defective item is $\left( \frac{29}{20} \right){{\left( \frac{19}{20} \right)}^{9}}$

4. Five cards are drawn successively with replacement from a well-shuffled deck of $52$cards. What is the probability that

(i) all the five cards are spades?

Ans: Let us have a notation K to be the random variable which measures the number of spade cards.

There are $13$spade cards and hece

$\text{ }p=\frac{13}{52}$

Therefore $q=1-\frac{1}{4}=\frac{3}{4}$

Clearly K follows Binomial Distribution with

$n=5,p=\frac{1}{4},q=\frac{3}{4}$

Probability of getting $5$cards spades is

$P\left( K=5 \right){{=}^{5}}{{C}_{5}}{{\left( \frac{3}{4} \right)}^{0}}{{\left( \frac{1}{4} \right)}^{5}}$

Hence we found that Probability of getting $5$cards spades is $\frac{1}{1024}$

(ii) only $3$ cards are spades?

Ans: Probability of only three cards are spades is given by

$P\left( K=3 \right){{=}^{5}}{{C}_{3}}{{\left( \frac{3}{4} \right)}^{2}}{{\left( \frac{1}{4} \right)}^{3}}$

Hence we found that Probability of only three cards are spades is $\frac{45}{512}$

(iii) none is a spade?

Ans: Probability of none of the cards is spade is given by

$P\left( K=0 \right){{=}^{5}}{{C}_{0}}{{\left( \frac{3}{4} \right)}^{5}}{{\left( \frac{1}{4} \right)}^{0}}$

Hence we found that Probability of only three cards are spades is $\frac{243}{1024}$

5. The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05$. What is the probability that out of 5 such bulbs

(i)none will fuse

Ans: Let us have a notation K to denote number of bulbs that will fuse after $150$ days in an experiment of $5$ trials

Given we have

$p=0.05$

$\therefore q=1-0.05=0.95$

Clearly K follows Binomial Distribution with

$n=5,p=0.05$

Probability that none of the bulbs will fuse is given by

$P\left( K=0 \right){{=}^{5}}{{C}_{0}}{{\left( 0.95 \right)}^{5}}{{\left( 0.05 \right)}^{0}}$

Hence probability that none of the bulbs will fuse is $0.7737$

(ii) not more than one will fuse

Ans: Probability that not more than one bulb will fuse is given by

$P\left( K\le 1 \right){{=}^{5}}{{C}_{1}}{{\left( 0.95 \right)}^{4}}{{\left( 0.05 \right)}^{1}}{{+}^{5}}{{C}_{0}}{{\left( 0.95 \right)}^{5}}{{\left( 0.05 \right)}^{0}}$

Hence Probability that not more than one bulb will fuse is $0.977$

(iii) ore than one will fuse

Ans: Probability that more than one bulb will fuse is given by

$P\left( K>1 \right)=1-P\left( K\le 1 \right)$

$\Rightarrow P\left( K>1 \right)=1-0.977$

Hence Probability that more than one bulb will fuse is $0.02$

(iv) At least one will fuse

Ans: Probability that at least one of the bulbs will fuse is given by

$P\left( K\ge 1 \right)=1-P\left( K<1 \right)$

$\Rightarrow P\left( K\ge 1 \right)=1-P\left( K=0 \right)$

$\Rightarrow P\left( K\ge 1 \right)=1-0.7737$

Hence Probability that at least one of the bulbs will fuse is $0.2263$

6. A bag consists of $10$ balls each marked with one of the digits $0$ to $9$. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit $0$?

Ans: Let us have K to denote the numbers of balls marked with $0$

Clearly k follows the binomial distribution with

$n=4,p=\frac{1}{10},q=\frac{9}{10}$

Probability that none of the numbers are marked with $0$ is given by

$P\left( K=0 \right){{=}^{4}}{{C}_{0}}{{\left( \frac{1}{10} \right)}^{5}}{{\left( \frac{9}{10} \right)}^{0}}$

$P\left( K=0 \right){{=}^{4}}{{C}_{0}}{{\left( \frac{1}{10} \right)}^{0}}{{\left( \frac{9}{10} \right)}^{4}}$

Hence Probability that none of the numbers are marked with $0$ is ${{\left( \frac{9}{10} \right)}^{4}}$

7. In an examination, $20$ s of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each . If the coin falls heads, he answers ‘true’; if it falls tail, he answers ‘false’. Find the probability that he answers at least $12$ s correctly.

Ans: Let us denote K to measure the number of correctly answered questions out of $20$

Given if a coin falls heads then he answered correctly and if it falls tails then he incorrectly answered

$\therefore p=\frac{1}{2}=q$

Clearly K follows Binomial Distribution with

$n=20,p=q=\frac{1}{2}$

Probability that he answers $12$ questions correctly is given by

$P\left( K\ge 12 \right)=P\left( K=12 \right)+P\left( K=13 \right)+P\left( K=14 \right)+...P\left( K=20 \right)$

$P\left( K\ge 12 \right)={{\left( \frac{1}{2} \right)}^{20}}\left[ ^{20}{{C}_{12}}{{+}^{20}}{{C}_{13}}{{+}^{20}}{{C}_{14}}+{{....}^{20}}{{C}_{20}} \right]$

Hence Probability that he answers $12$ questions correctly is ${{\left( \frac{1}{2} \right)}^{20}}\left[ ^{20}{{C}_{12}}{{+}^{20}}{{C}_{13}}{{+}^{20}}{{C}_{14}}+{{....}^{20}}{{C}_{20}} \right]$

8. Suppose X has a binomial distribution B$\left( 6,\frac{1}{2} \right)$. Show that $X=3$is the most likely outcome.

(Hint: P(X = 3) is the maximum among all P(xi), xi = 0, 1, 2, 3, 4, 5, 6)

Ans: Given X follows binomial distribution with

$n=6,p=\frac{1}{2}$

\[\Rightarrow P\left( X=x \right){{=}^{6}}{{C}_{x}}{{\left( \frac{1}{2} \right)}^{x}}{{\left( \frac{1}{2} \right)}^{6-x}}\]

\[\Rightarrow P\left( X=x \right){{=}^{6}}{{C}_{x}}{{\left( \frac{1}{2} \right)}^{6}}\]

Hence $P\left( X=x \right)$ will be maximum when $^{6}{{C}_{x}}$will be maximum

And since $n=6$ is an even number therefore $^{6}{{C}_{x}}$is maximum at $x=3$

Hence $P\left( X=3 \right)$ will be maximum

9. On a multiple choice examination with three possible answers for each of the five s, what is the probability that a candidate would get four or more correct answers just by guessing?

Ans: Let us denote K to measure the numbers of correct answers by guessing

Probability of getting a correct answer is $p=\frac{1}{3}$

Probability of getting a non correct answer is $q=\frac{2}{3}$

Clearly K follows binomial distribution with

$n=5,p=\frac{1}{3}$

Probability of guessing more than correct answers is given by

$P\left( K\ge 4 \right){{=}^{5}}{{C}_{4}}{{\left( \frac{1}{3} \right)}^{4}}\left( \frac{2}{3} \right){{+}^{5}}{{C}_{5}}{{\left( \frac{1}{3} \right)}^{5}}$

Hence Probability of guessing more than correct answers is $\frac{11}{243}$

10. A Person buys a lottery ticket in $50$ lotteries, in each of which his change of winning a prize is $\frac{1}{100}$ . What is the probability that he will in a prize

At least once

Ans: Let K be a random variable that represents the number of winning prizes

It is visible that K follows Binomial Distribution with

$n=50,p=\frac{1}{100}$

$\therefore q=\frac{99}{100}$

Probability of winning at least once is given by

$P\left( K\ge 1 \right)=1-P\left( K<1 \right)$

$P\left( K<1 \right){{=}^{50}}{{C}_{0}}{{\left( \frac{1}{100} \right)}^{0}}{{\left( \frac{99}{100} \right)}^{50}}$

Hence probability of winning at least once is $1-\left( \frac{99}{100} \right)$

Exactly once

Ans: Probability of winning exactly once is given by

$P\left( K=1 \right){{=}^{50}}{{C}_{1}}{{\left( \frac{1}{100} \right)}^{1}}{{\left( \frac{99}{100} \right)}^{49}}$

Hence probability of winning exactly once is $\frac{1}{2}{{\left( \frac{99}{100} \right)}^{49}}$

At least twice?

Ans: Probability of at least twice is given by

$P\left( K\ge 2 \right)=1-P\left( K\le 1 \right)$

Now $P\left( K\le 1 \right)=1-\left( \frac{99}{100} \right)+\frac{1}{2}{{\left( \frac{99}{100} \right)}^{49}}$

Hence Probability of at least twice is $1-\frac{149}{100}{{\left( \frac{99}{100} \right)}^{49}}$

11. Find the probability of getting $5$exactly twice in $7$ throws of a die.

Ans: Let K be the random variable that represents number of outcome to be $5$

Clearly K follows Binomial Distribution with

$n=7,p=\frac{1}{6},q=\frac{5}{6}$

Probability of getting $5$ exactly twice is given by

$P\left( K=2 \right){{=}^{7}}{{C}_{2}}{{\left( \frac{1}{6} \right)}^{2}}{{\left( \frac{5}{6} \right)}^{5}}$

Hence we found that Probability of getting $5$ exactly twice is $\frac{7}{12}\times {{\left( \frac{5}{6} \right)}^{5}}$

12. Find the probability of throwing at most $2$ sixes in $6$ throws of a single die.

Ans: Let K be the random variable that represents the number of sixes.

Clearly K follows the Binomial Distribution with

$n=6,p=\frac{1}{6},q=\frac{5}{6}$

Probability of at most two sixes is given by

$\text{ }P\left( K\le 2 \right)=P\left( K=0 \right)+P\left( K=1 \right)+P\left( K=2 \right)$……($1$)

$\text{ }P\left( K=0 \right){{=}^{6}}{{C}_{0}}{{\left( \frac{1}{6} \right)}^{0}}{{\left( \frac{5}{6} \right)}^{6}}$

$\Rightarrow P\left( K=0 \right)={{\left( \frac{5}{6} \right)}^{6}}$

$\text{ }P\left( K=1 \right){{=}^{6}}{{C}_{1}}{{\left( \frac{1}{6} \right)}^{1}}{{\left( \frac{5}{6} \right)}^{5}}$

$\Rightarrow P\left( K=1 \right)={{\left( \frac{5}{6} \right)}^{5}}$

\[\text{ }P\left( K=2 \right){{=}^{6}}{{C}_{2}}{{\left( \frac{1}{6} \right)}^{2}}{{\left( \frac{5}{6} \right)}^{4}}\]

$\Rightarrow P\left( K=2 \right)=\frac{6\times 5}{2\times 36}{{\left( \frac{5}{6} \right)}^{4}}$

$\Rightarrow P\left( K=2 \right)=\frac{5}{24}{{\left( \frac{5}{6} \right)}^{4}}$

Put all these in equation ($1$) as shown

$\text{ }P\left( K\le 2 \right)={{\left( \frac{5}{6} \right)}^{6}}+{{\left( \frac{5}{6} \right)}^{5}}+\frac{5}{12}{{\left( \frac{5}{6} \right)}^{4}}$

$\Rightarrow P\left( K\le 2 \right)={{\left( \frac{5}{6} \right)}^{4}}\left( \frac{25}{36}+\frac{5}{6}+\frac{5}{12} \right)$

$\Rightarrow P\left( K\le 2 \right)={{\left( \frac{5}{6} \right)}^{4}}\left( \frac{35}{18} \right)$

Hence Probability of at most two sixes is $0.9377$

13. It is known that $10$% of certain articles manufactured are defective. What is the probability that in a random sample of $12$ such articles, $9$ are defective?

Ans: Let K be the random variable that represents the number of selecting defective items

It is visible that K follows binomial distribution with

$n=12,p=0.1,q=0.9$

Probability of selecting $9$ defective articles is given by

$P\left( K=9 \right){{=}^{12}}{{C}_{9}}{{\left( 0.1 \right)}^{9}}{{\left( 0.9 \right)}^{3}}$

$\Rightarrow P\left( K=9 \right)=220\times \frac{{{9}^{3}}}{{{10}^{12}}}$

Hence Probability of selecting $9$ defective articles is $22\times \frac{{{9}^{3}}}{{{10}^{11}}}$

14. In a box containing $100$ bulbs, $10$ are defective. The probability that out of a sample of $5$ bulbs, none is defective is?

Ans: Let K be the random variable that represents number of defective bulbs

It is visible that K follows Binomial Distribution with

$n=5,p=0.1,q=0.9$

Probability of getting no defective bulbs is given by

$P\left( K=0 \right){{=}^{5}}{{C}_{0}}{{\left( 0.1 \right)}^{0}}{{\left( 0.9 \right)}^{5}}$

Hence Probability of getting no defective bulbs is ${{\left( \frac{9}{10} \right)}^{5}}$

15: The probability that a student is not swimming is $\frac{1}{5}$ . The probability that out of five students, four are swimmers is?

Ans: Let K be the random variable that represents number of swimmers

It is easily visible that K follows binomial Distribution with

$n=5,p=\frac{1}{5},q=\frac{4}{5}$

Probability of four students to be swimmers is given by

$P\left( K=4 \right){{=}^{5}}{{C}_{4}}{{\left( \frac{1}{5} \right)}^{4}}\frac{4}{5}$

Hence Probability of four students to be swimmers is $^{5}{{C}_{4}}{{\left( \frac{1}{5} \right)}^{4}}\frac{4}{5}$

### Miscellaneous

1: If A and B are two events such that $P\left( A \right)\ne 0$. Find $P\left( B\left| A \right. \right)$ if

(i) A is a subset of B

Ans: It is given in the question that $P\left( A \right)\ne 0$ and A is a subset of B

$\therefore P\left( A\bigcap B \right)=P\left( A \right)$

$\therefore P\left( B\left| A \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}$

$\Rightarrow P\left( B\left| A \right. \right)=\frac{P\left( A \right)}{P\left( A \right)}=1$

(ii) $A\bigcap B=\phi $

Ans: since it is given that $A\bigcap B=\phi $

Therefore $\text{ }P\left( A\bigcap B \right)=0$

$\therefore P\left( B\left| A \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}=0$

2. A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

Ans: The sample space for a family to have two children is given by

$S=\left\{ BB,GG,BG,GB \right\}$

Let us have following notations and their probabilities given as shown

N: Both children are males

$P\left( N \right)=\frac{1}{4}$

K: at least one of the children is male

\[P\left( K \right)=\frac{3}{4}\]

$\because N\bigcap K=BB$

$\therefore P\left( N\bigcap K \right)=\frac{1}{4}$

Now probability that both of the children are males provided at least one of the child is male is given by

$P\left( N\left| K \right. \right)=\frac{P\left( N\bigcap K \right)}{P\left( K \right)}$

$P\left( N\left| K \right. \right)=\frac{\frac{1}{4}}{\frac{3}{4}}$

Hence probability that both of the children are males provided at least one of the child is male is $\frac{1}{3}$

(ii)Find the probability that both children are females, if it is known that the elder child is a female.

Ans: The sample space for a family to have two children is given by

$S=\left\{ BB,GG,BG,GB \right\}$

Let us have following notations and their probabilities given as shown

N: Both children are females

$P\left( N \right)=\frac{1}{4}$

K: elder child is a female

\[P\left( K \right)=\frac{2}{4}\]

$\because N\bigcap K=GG$

$\therefore P\left( N\bigcap K \right)=\frac{1}{4}$

Now probability that both of the children are males provided atleast one of the child is male is given by

$P\left( N\left| K \right. \right)=\frac{P\left( N\bigcap K \right)}{P\left( K \right)}$

$P\left( N\left| K \right. \right)=\frac{\frac{1}{4}}{\frac{2}{4}}$

Hence probability that both of the children are males provided atleast one of the child is male is $\frac{1}{2}$

3. Suppose that $5$% of men and $0.25$% of women have grey hair. A haired person is selected at random. What is the probability of this person being male? Assume that there are equal numbers of males and females.

Ans: It is given that $5$percent of males and $0.25$ percent of females have grey hair

Therefore total people having grey hair is $\left( 5+0.25 \right)=5.25$ percent

Hence probability of male being haired is $\frac{5}{5.25}$

4. Suppose that $90$ % of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Ans: Given probability of a person being right-handed is$p=0.9$

Since a person can only be right-handed or left handed

Therefore it follows binomial distribution with

$n=10,p=\frac{9}{10},q=\frac{1}{10}$

Probability that at least $6$ people are right handed is given by

\[{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \frac{9}{10} \right)}}^{k}}{{\left( \frac{1}{10} \right)}^{10-k}}\]

Therefore probability that at most $6$ people are right handed is given by

\[1-{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \frac{9}{10} \right)}}^{k}}{{\left( \frac{1}{10} \right)}^{10-k}}\]

5. An urn contains$25$balls of which $10$ balls bear a mark ‘X’ and the remaining $15$ bear a mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If $6$ balls are drawn in this way, find the probability that

(i)all will bear ‘X’ mark.

Ans: from the given data we have

Probability of balls bearing X mark is $p=\frac{10}{25}$

Probability of balls bearing Y mark is $q=\frac{15}{25}$

Hence we can say that random variable K which measures the number of Y marked balls follows binomial distribution with

$n=6,p=\frac{2}{5},q=\frac{3}{5}$

Probability that all ball bear X is same as none of balls bear Y which is given by

$P\left( K=0 \right){{=}^{6}}{{C}_{0}}{{\left( \frac{3}{5} \right)}^{0}}{{\left( \frac{2}{5} \right)}^{6}}$

Hence Probability that all ball bear X is ${{\left( \frac{2}{5} \right)}^{6}}$

(ii) not more than 2 will bear the ‘Y’ mark.

Ans: Probability of not more than $2$MY balls are selected is given by

$\text{ }P\left( k\le 2 \right)=P\left( K=0 \right)+P\left( K=1 \right)+P\left( K=2 \right)$

$\Rightarrow P\left( k\le 2 \right){{=}^{6}}{{C}_{0}}{{\left( \frac{3}{5} \right)}^{0}}{{\left( \frac{2}{5} \right)}^{6}}{{+}^{6}}{{C}_{1}}{{\left( \frac{3}{5} \right)}^{1}}{{\left( \frac{2}{5} \right)}^{5}}{{+}^{6}}{{C}_{2}}{{\left( \frac{3}{5} \right)}^{2}}{{\left( \frac{2}{5} \right)}^{4}}$

$\Rightarrow P\left( k\le 2 \right)={{\left( \frac{2}{5} \right)}^{4}}\left( \frac{4}{25}+6{{\left( \frac{3}{5} \right)}^{1}}\left( \frac{2}{5} \right)+15{{\left( \frac{3}{5} \right)}^{2}} \right)$

Hence Probability of not more than $2$ balls are selected is $7{{\left( \frac{2}{5} \right)}^{4}}$

(iii) At least one ball will bear the ‘Y’ mark.

Ans: Probability of at least one ball bear Y is given by

$P\left( K\ge 1 \right)=1-P\left( k=0 \right)$

$\Rightarrow P\left( K\ge 1 \right)=1-{{\left( \frac{2}{5} \right)}^{6}}$

Hence Probability of at least one ball bear Y is $1-{{\left( \frac{2}{5} \right)}^{6}}$

(iv) The number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Ans: Since there were $6$ balls therefore each are of $3$ if equal

Probability that number of X and Y is equal is given by

\[P\left( K=3 \right){{=}^{6}}{{C}_{3}}{{\left( \frac{2}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{3}}\]

Hence Probability that number of X and Y is equal is $\frac{834}{3125}$

6. In a hurdle race, a player has to cross $10$ hurdles. The probability that he will clear each hurdle Is $\frac{5}{6}$ . What is the probability that he will knock down fewer than $2$ hurdles?

Ans: Given probability that he will clear each hurdle is $p=\frac{5}{6}$

And hence probability that he knocks down is given as $q=\frac{1}{6}$

Let k be the random variable that represents the hurdles he knocks down

Clearly K follows binomial distribution with

$n=10,p=\frac{5}{6},q=\frac{1}{6}$

Probability that he knocks down less than $2$hurdles is given by

$P\left( K=0 \right)+P\left( K=1 \right)$

$^{10}{{C}_{0}}{{\left( \frac{1}{6} \right)}^{0}}{{\left( \frac{5}{6} \right)}^{10}}{{+}^{10}}{{C}_{1}}{{\left( \frac{1}{6} \right)}^{1}}{{\left( \frac{5}{6} \right)}^{9}}=\frac{{{5}^{10}}}{{{6}^{9}}\times 2}$

Hence Probability that he knocks down less than $2$hurdles is $\frac{{{5}^{10}}}{{{6}^{9}}\times 2}$

7. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Ans: From the question we have

Probability of getting a six is $\text{ }p=\frac{1}{6}$

Probability of getting no six is $q=\frac{5}{6}$

Let K be the random variable that represents the number of sixes

Clearly K follows binomial distribution with

$n=5,p=\frac{1}{6},q=\frac{5}{6}$

Probability that two sixes come in first two throws is given by

$P\left( K=2 \right){{=}^{5}}{{C}_{2}}{{\left( \frac{1}{6} \right)}^{2}}{{\left( \frac{5}{6} \right)}^{3}}$

Also Probability that third sixe come is given by

$\frac{1}{6}P\left( K=2 \right)$

Hence probability of getting three sixes is $\frac{10\times 125}{{{6}^{6}}}$

8: If a leap year is selected at random, what is the chance that it will contain $53$Tuesdays?

Ans: In a leap year, we have $366$ days i.e., $52$weeks and $2$ days.

In $52$ weeks, we have $52$ Tuesdays.

Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining $2$ days will be Tuesdays.

The remaining $2$days can be any of the following

Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday and Sunday and Monday

Total number of cases = 7

⸫ Probability that a leap year will have $53$ Tuesdays $=\frac{2}{7}$

9. An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

Ans: Given that the probability of success is twice that of failure

$\therefore p=2q$

Also we know that sum of probabilities of exhaustive events equals $1$

$\therefore p=\frac{2}{3},q=\frac{1}{3}$

Let K be the random variable that represents successes

$\therefore P\left( K\ge 4 \right)=P\left( K=4 \right)+P\left( K=5 \right)+P\left( K=6 \right)$

$\Rightarrow P\left( K\ge 4 \right){{=}^{6}}{{C}_{4}}{{\left( \frac{2}{3} \right)}^{4}}{{\left( \frac{1}{3} \right)}^{2}}{{+}^{6}}{{C}_{3}}{{\left( \frac{2}{3} \right)}^{3}}{{\left( \frac{1}{3} \right)}^{1}}{{+}^{6}}{{C}_{6}}{{\left( \frac{2}{3} \right)}^{6}}{{\left( \frac{1}{3} \right)}^{0}}$

Thus on solving we found the probability of getting at least $4$ successes in six trials is $\frac{31}{9}{{\left( \frac{2}{3} \right)}^{4}}$

10. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Ans: Let a person tosses the coin n times , n being Bernoulli Trials

Here probability of head equals that of tails and each equal to $\frac{1}{2}$

Numbers of trials for which at least one head is more than $0.9$ is given by

$1-P\left( K=0 \right)>0.9$

$\Rightarrow 1{{-}^{n}}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{n}}>\frac{9}{10}$

$\Rightarrow {{2}^{n}}>10$

Hence the person must toss more than $4$ times

11. In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

Ans: We have

Probability of getting a six is $p=\frac{1}{6}$

Probability of getting no six is $q=\frac{5}{6}$

There are three possible ways to get a six in three trials:

He gets a six only at first throw

probability $=\frac{1}{6}$

Rupee won $=1$

He gets a six at second throw

probability $=\frac{1}{6}\times \frac{5}{6}$

Rupee won $=-1+1=0$

He gets a six at third trial

probability $=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{25}{216}$

Rupee won $=-1-1+1=-1$

Thus Expected value can be calculated as shown

$1\times \frac{1}{6}+0\times \frac{5}{36}+-1\times \frac{25}{216}=\frac{11}{216}$

12. Suppose we have four boxes. A, B, C and D containing coloured marbles as given below.

Box | Red | White | Black |

A | 1 | 6 | 3 |

B | 6 | 2 | 2 |

C | 8 | 1 | 1 |

D | 0 | 6 | 4 |

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Ans: Let us have following notations

R be the event of drawing red marble

E be the event of selecting the box A

F be the event of selecting the box B

G be the event of selecting the box C

Total numbers of marbles is $40$

Number of red marbles is $15$

$\therefore P\left( R \right)=\frac{15}{40}$

Number of red marbles in box A i.e $n\left( R\bigcap E \right)=1$

Number of red marbles in box B i.e $n\left( R\bigcap F \right)=6$

Number of red marbles in box C i.e $n\left( R\bigcap G \right)=8$

Now the probability that red marble is picked from box A is given by

$P\left( R\left| E \right. \right)=\frac{P\left( R\bigcap E \right)}{P\left( E \right)}$

$\Rightarrow P\left( R\left| E \right. \right)=\frac{\frac{1}{40}}{\frac{15}{40}}$

Hence E be the event of selecting the box A is $\frac{1}{15}$

Similarly, probability that red marble is picked from box B is given by

$P\left( R\left| F \right. \right)=\frac{\frac{6}{40}}{\frac{15}{40}}$

Hence probability that red marble is picked from box B is $\frac{2}{5}$

Similarly, probability that red marble is picked from box C is given by

$P\left( R\left| G \right. \right)=\frac{\frac{8}{40}}{\frac{15}{40}}$

Hence probability that red marble is picked from box C is $\frac{8}{15}$

13: Assume that the chances of the patient having a heart attack are $40$%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30$% and prescription of certain drugs reduces its changes by $25$%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Ans: Let us have following notations

E: The events when the person took yoga and meditation courses

$\therefore P\left( E \right)=\frac{1}{2}$

F:The events when the person took drugs

$\therefore P\left( F \right)=\frac{1}{2}$

G: the person suffered heart attack

$\therefore P\left( G \right)=0.4$

From the question also we have

$P\left( G\left| E \right. \right)=0.4\times 0.7=0.28$

$P\left( G\left| F \right. \right)=0.4\times 0.75=0.30$

Now probability that found person has heart attack despite of having yoga and meditation courses is given by

$P\left( E\left| G \right. \right)=\frac{P\left( E \right)\times P\left( G\left| E \right. \right)}{P\left( E \right)\times P\left( G\left| E \right. \right)+P\left( F \right)\times P\left( G\left| F \right. \right)}$

$P\left( E\left| G \right. \right)=\frac{0.5\times 0.28}{P0.5\times 0.28+0.5\times 0.3}$

Hence probability that found person has heart attack despite of having yoga and meditation courses is $\frac{14}{29}$

14. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$)

Ans: It is clear that total number of determinants of second order entries being $0's\text{ or 1 }\!\!'\!\!\text{ s}$ is ${{2}^{4}}$

The value of determinant is positive for cases as shown

$\left. \left| \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right. \right|,\left. \left| \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right. \right|,\left. \left| \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right. \right|$

i.e favourable case is $3$

Hence probability that value of determinant is positive is $\frac{3}{16}$.

15: An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P (A fails) = $0.2$

P (B fails alone) = $0.15$

P (A and B fails) = $0.15$

Evaluate the following probabilities

(i) P (A fails | B has failed)

Ans: Let us have following notations

E: A fails

F: B fails

Given in the question

$P\left( E \right)=0.2$

$P\left( E\bigcap F \right)=0.15$

$P\left( E'\bigcap F \right)=0.15$

We know that

$P\left( E'\bigcap F \right)=P\left( E \right)-P\left( E\bigcap F \right)$

$\Rightarrow P\left( E \right)=0.3$

Now the probability that A fails given B has failed is given by

$P\left( E\left| F \right. \right)=\frac{0.15}{0.3}$

Hence the probability that A fails given B has failed is $0.5$

(ii) P (A fails alone)

Ans: Probability that A fails alone is given by

$P\left( E\bigcap F' \right)=P\left( F \right)-P\left( E\bigcap F \right)$

$P\left( E\bigcap F' \right)=0.05$

Hence Probability that A fails alone is $0.05$

16. Bag I contains $3$ red and $4$ black balls and Bag II contains $4$ red and $5$ black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Ans: Let us have following notations that

E: red ball is transferred

$P\left( E \right)=\frac{3}{7}$

F: black ball is transferred

$P\left( F \right)=\frac{4}{7}$

G: red ball is drawn

When a red ball is transferred

$P\left( G\left| E \right. \right)=\frac{5}{10}$

Similarly, When a black ball is transferred

$P\left( G\left| F \right. \right)=\frac{4}{10}$

$P\left( F\left| G \right. \right)=\frac{\frac{4}{7}\times \frac{4}{10}}{\frac{4}{7}\times \frac{4}{10}+\frac{3}{7}\times \frac{5}{10}}$

Hence probability that the transferred ball is black is $\frac{16}{31}$

17. If A and B are two events such that $P\left( A \right)\ne 0$and $P\left( B\left| A \right. \right)=1$, then what result can follow?

Ans: Given in the question $P\left( A \right)\ne 0$ and $P\left( B\left| A \right. \right)=1$

And we know that

$P\left( B\left| A \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}$

$\Rightarrow P\left( A \right)=P\left( A\bigcap B \right)$

Hence $A\subset B$

18. If $P\left( A\left| B \right. \right)>P\left( A \right)$), then what is correct?

Ans: Given $P\left( A\left| B \right. \right)>P\left( A \right)$

And we know that

$P\left( A\left| B \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$

$\Rightarrow \frac{P\left( A\bigcap B \right)}{P\left( B \right)}>P\left( A \right)$

Hence we found that $P\left( B\left| A \right. \right)>P\left( B \right)$

19. If A and B are any two events such that $P\left( A \right)+P\left( B \right)-P\left( A\text{ and B} \right)=P\left( A \right)$then what result can follow?

Ans: Given in the question we have

$P\left( A \right)+P\left( B \right)-P\left( A\bigcap \text{B} \right)=P\left( A \right)$

$\Rightarrow P\left( B \right)=P\left( A\bigcap B \right)$

Hence

$P\left( A\left| B \right. \right)=1$

### NCERT Solutions for Class 12 Maths – Free PDF Download

You can now get the Probability Class 12 NCERT Book Solutions PDF Download for free on Vedantu, so you can refer to it whenever you wish to. The Probability Class 12 PDF solutions are handy, which means that whenever you have any doubts on any topic on probability, you can refer to the solutions to get your concepts cleared. Every solution is solved in detailed steps to explain the concept well to you.

### Chapter 13 – Probability

### NCERT Solutions for Class 12 Maths Chapter 13

#### 13.1 Introduction

In the earlier Class 12 Maths probability chapters, you have already learnt about what probability is. In any random event, the measure of uncertainty is measured by probability. A. N. Kolmogorov had formulated the axiomatic approach which was studied in the chapters. He treated probability as a function of the outcome of any experiment. In the case where there are equally likely outcomes, the relation between classical and the axiomatic theory has also been discussed. This lets you obtain the probability in the case of samples that have a discrete sample size. The addition rule in probability has also been studied.

Probability Class 12th chapter will explain the concepts of the conditional probability of any event given that the other event has occurred. This helps in understating the Bayes’ theorem, independence of events and the multiplication rule in probability. The probability distribution of random variables and the mean and the variance of probability distribution will also be studied. The binomial distribution is referred to as a discrete probability distribution which will be explained in this chapter. There are activities in the chapter that help to make the concepts clearer. The experiments are conducted assuming that there are outcomes that are equally likely.

#### 13.2 Conditional Probability

In this Class 12 Probability Solutions, you learn how if there are two events that come from the same sample space, the occurrence of one of the events affects the occurrence of the other? The conditional probability of event B is the probability that the event will take place given that you already have knowledge that event A has already taken place. The probability notation is given by P(B|A) which means the probability of B given A.

In this case where the two events A and B are independent where the event A does not affect the probability of event B then the conditional probability of the event B given the event A is P(B).

However, if the two events A and B are not independent, the probability of the intersection of A and B that is the probability of both the events occurring is denoted by

P(A and B) = P(A)P(B|A).

This can help you to get the probability of P(B|A) which is obtained by

P(B|A) = P(A∩B)/P(A)

The expression is valid when P(A) is greater than 0.

Thus if A and B are two events in a sample space say S, the conditional probability of

P(A|B) = P(A∩B)/P(B)

Where P(B)>0

#### 13.2.1 Properties of Conditional Probability

Property 1: If E and F are the events of a sample space say S, P(S|F) = P(F|F) = 1

Property 2: If A and B are two events in a sample space S and F is an event of S such that P(F) ≠ 0, P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F).

Property 3: P(A′|B) = 1 − P(A|B)

#### 13.3 Multiplication theorem on probability

This Probability Class 12 section explains the multiplication theorem of probability. You already know that the conditional probability of event A given that B has occurred is denoted by P(A|B) and is given by:

P(A|B) = P(A∩B)P(B)

Where P(B) is not equal to 0

Thus, P(A∩B) = P(B)×P(A|B)

P(B|A) = P(B∩A)P(A)

Where P(A) is not equal to 0.

Thus, P(B∩A) = P(A)×P(B|A)

Now Since, P(A∩B) = P(B∩A)

P(A∩B) = P(A)×P(B|A)

P(A∩B) = P(B)×P(A|B) = P(A)×P(B|A) where,

P(A) and P(B) is not equal to 0

The above result is known as the multiplication rule of probability.

For independent events A and B, P(B|A) = P(B). The equation then can be modified into,

P(A∩B) = P(B) × P(A)

#### 13.4 Independent Events

The NCERT Solutions Class 12 Probability also explains independent events which are those events that when occurs does not affect any other event. Like if a coin is flipped in the air and the outcome is ahead. If you flip the coin again, the outcome is a tail. In both cases, the occurrence of each event is independent of the other.

The set of outcomes is known as events and this is explained well in the Probability Chapter Class 12 PDF. If the probability of an outcome of an event say A is not affected by the probability of occurrence of another event B, it is said that A and B are independent events.

#### 13.5 Bayes’ Theorem

In the NCERT Solutions for Class 12 Maths Chapter 13 Probability PDF section, you learn about what Bayes’ theorem is. Bayes theorem describes the probability of an event occurring that is related to any condition. This is also considered in the case of conditional probability. Like for example, suppose you have a bag that contains three balls of different colours, say black, blue and red. You need to calculate the probability of taking out a blue ball from the bag out of the three balls. Here, the probability of the event occurring is calculated depending on the other conditions which are known as conditional probability.

#### 13.5.1 Partition of a Sample Space – H3

This Probability NCERT Solutions Class 12 section explains what partition of sample space is. Set of events say E1, E2, …, En represents the partition of sample space S is if

The events E1, E2, …, En represent a partition of the sample space S if they are pairwise disjoint, are exhaustive and have nonzero probabilities.

#### 13.5.2 Theorem of Total Probability

Suppose there are events, say C1, C2 . . . Cn and they form partitions of the sample space S, where all the events have a non-zero probability of occurrence. Then for any event, A associated with S, according to the total probability theorem,

P(A) = ∑k=0nP(Ck)P(A|Ck)

#### 13.6 Random Variables and Its Probability Distribution

#### Random Variables

A random variable is a variable that is capable of changing its value. The value will differ based on the varied outcomes of any experiment. If the value of any variable is dependent on the outcome from a random experiment, it is called a random variable. The random variable can take any real value.

A random variable is a real-valued function. Its domain is a sample space S of a random experiment. A random variable is capable of assigning a numerical value to every outcome in a sample space. The random variable can either be discrete or continuous. A discrete random variable is the one that assumes only specific values in an interval. Like 1.2,3… Otherwise, the random variable is a continuous random variable.

#### 13.6.1 Probability Distribution of a Random Variable

The probability distribution of a random value is what defines the probability of its value that is unknown. The random variable could either be discrete or continuous. They can also be both.

There can be two random variables with an equal probability distribution and yet they can vary with how they are in respect to the other random variables or whether they are independent of the random variables. The recognition of the outcome of the value that is randomly chosen as per the probability distribution of the variable is the random variate.

For any event of any random experiment, it is possible to find its corresponding probability. Even for the different random variable values, it is possible to find its respective probability. The probability distribution of the random variable is the value of the random variable along with its corresponding probabilities. Properties of Probability Distribution

The probability distribution of a random variable is explained in Class 12 Maths Ch 13 NCERT solutions where X is P(X = xi) = pi for x = xi and P(X = xi) = 0 for x ≠ xi.

The range of probability distribution for all possible values of a random variable is from 0 to 1, i.e., 0 ≤ p(x) ≤ 1.

#### 13.6.2 Mean of a random variable

If X is a random variable and has the possible valued like x1,x2,x3,…,xn occurring with probabilities p1,p2,p3,…,pn, respectively, the mean of the random variable denoted by μ is the weighted average of the possible values of X. Each of the values is weighted by the probability with which it occurs.

The mean of a random variable can also be said as the expectation of X.

E(X) = μ = ∑i=1n xipi

=x1p1 + x2p2 + ⋯ + xnpn

#### 13.6.3 Variance of a Random Variable

The variance tells about how spread out the X values is around its mean. The variance of a random variable is denoted by σ2x with the values of X as x1,x2,x3,…,xn and their occurrence with the probability of p1,p2,p3,…,pn.

Var(X) = σ2x = ∑i=1n(xi − μ)2pi

Var(X) = ∑i=1n (xi)2pi + ∑i=1n μ2pi − ∑o=1n 2xiμpi

Var(X) = ∑i=1n (xi)2pi + μ2 ∑ ni=1 pi − 2μ∑i=1n xipi

Here,

∑i=1nxipi = μ (Mean of X) and ∑i=1n pi = 1 (sum of probabilities of all the outcomes of an event is 1). Substituting the values, we get

Var(X) = ∑i=1n(xi)2pi + μ2 − 2μ2

σ2x = Var(X) = ∑i=1n (xi)2pi − μ2

Var(X) = E(X2) – [E(X)]2

Where,

E(X2) = ∑i=1n(xi)2pi and E(X) = ∑i=1nxipi

#### 13.7 Bernoulli Trials and Binomial Distribution

This Ch 13 Maths Class 12 section explains Bernoulli trials and the binomial distribution in brief. Bernoulli trial or binomial trial is where there are only two possible outcomes. In the case of the binomial distribution, we can get the number of successes in the sequence of experiments that are independently conducted.

#### 13.7.1 Bernoulli Trials

There are many random experiments that can have only two of the possible outcomes. These are success and failure. Like for example, the toss of a coin which can be heads or tails or the result that something can either be defective or not defective.

These are independent trials and can have only two of the possible outcomes. These are known as the Bernoulli trials. Here are the conditions of a trial to be categorized into a Bernoulli trial.

The number of trials should be finite

The trials should be independent

Each trial should have only two outcomes with are success and failure

The probability of either success or failure will not change for every trail

#### 13.7.2 Binomial Distribution

Suppose there is a random experiment that can exactly have only two outcomes and this is repeated for n number of times and independently. The probability of success is denoted by p and the probability of failure is denoted by q.

Out of these trials, we get success in x number of trials and we get failure in the remaining (n-x) number of trials. Thus the number of trials in which we can have success is nCx

A random variable has a binomial distribution when

P(X = x) = p(x) = nCx px qn-x,

for x = 0, 1, … , n and P(X = x) = 0 otherwise. Here, q = 1 – p. The random variable X is a binomial variable. Thus you can say that a binomial trial is a set of Bernoulli trials which has n independent trials.

### Here are the Conditions for a Binomial Distribution.

Each trial should result in only two outcomes which are success and failure

The number of trials says n is finite

Each trial is independent of each other

The probability of success which is p and the probability of failure which is q stay constant in each trial.

### Key Features of the NCERT Solution for Class 12 Maths Chapter 13

Probability is an important topic which will be studied further if you take up mathematics at the undergraduate level. This is why it is important to build on your concepts and clear your doubts right from the start. This is why there are Vedantu’s expert teachers who have solved the NCERT probability Class 12 questions with a complete explanation to ensure that you do not just learn the concept for your examination but also master the topic.

The Ch 13 Class 12 Maths NCERT solution by Vedantu prepares you well for your examination.

The solutions are clear and detailed to give you a thorough explanation of the topic

It is important to keep referring to these solutions to understand the various approaches to the questions.

Solving these questions and checking out the solutions will make you more confident in attempting the probability question in the examination.

### Chapter wise NCERT Solutions for Class 12 Maths

### Access Other Exercises of Class 12 Maths Chapter 13

Chapter 13 Probability All Exercises in PDF Format | |

3 Questions and Solutions | |

18 Questions and Solutions | |

14 Questions and Solutions | |

17 Questions and Solutions | |

15 Questions and Solutions |

## FAQs on NCERT Solutions Class 12 Maths Chapter 13 Probability

**1. How many topics and sub-topics are there in this chapter?**

Studying Maths chapter 13 of Class 12 namely Probability helps the students to understand various relevant concepts such as Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution etc.

Following concepts are covered in Chapter 13 of NCERT Solutions for Class 12. Take a look at the following list:

13.1 Introduction

13.2 Conditional Probability

13.2.1 Properties of conditional probability

13.3 Multiplication Theorem on Probability

13.4 Independent Events

13.5 Bayes’ Theorem

13.5.1 Partition of a sample space

13.5.2 Theorem of total probability

13.6 Random Variables and its Probability Distributions

13.6.1 Probability distribution of a random variable

13.6.2 Mean of a random variable

13.6.3 Variance of a random variable

13.7 Bernoulli Trials and Binomial Distribution

13.7.1 Bernoulli trials

13.7.2 Binomial distribution

**2. What is Conditional probability?**

Conditional probability basically depicts the probability of an event taking place with some relationship to one or more events. Suppose E and F are two events associated with the same sample space of a random experiment, then the conditional probability of the event E will be under the condition of which the event F has occurred, written as P (E | F).

**3. How many exercises are there in Class 12 Maths Chapter 13?**

Chapter 13 Probability NCERT Solutions contains several exercises. Every exercise contains the answer to each question that has been asked in every exercise of chapter 13. All the answers are accurate and provided with step-by-step instructions. This is considered as the most helpful study material for every student to do their home assignments as well as practice various sample papers. Here’s the number of questions of all the exercises -

Chapter 13 Exercises 13.1 - 17 questions

Chapter 13 Exercises 13.2 - 18 questions

Chapter 13 Exercises 13.3 - 14 questions

Chapter 13 Exercises 13.4 - 17 questions

Chapter 13 Exercises 13.5 - 15 questions

**4. Why Vedantu’s NCERT Solutions for Class 12 Maths Chapter 13 is beneficial for the students?**

NCERT Solutions for Class 12 Maths Chapter 13 Probability is provided in PDF format on our site which you can download very easily as per your convenience. These solutions are designed and prepared by the best teachers and subject matter experts from the respective fields. All the important topics are covered in every answer to the questions from the exercises of the NCERT textbook and each answer comes with a detailed step-by-step explanation to help students understand concepts better. These NCERT solutions play a crucial role in building the foundation stronger and also assist in your preparation for other competitive exams such as JEE Main, JEE Advanced, Olympiad.

Chapter 13 Probability of the NCERT Solutions for Class 12 Maths covers multiple exercises which are given in between and at the end of the chapter. Check our website or download our mobile app to access the NCERT Solutions in PDF downloadable format.

**5. What all things are added in NCERT Solutions for Chapter 13 Probability of Class 12 Maths?**

NCERT Solutions Chapter 13 Probability of Class 12 Maths teaches all of the fundamental concepts connected to the conditional probability that is useful in comprehending the Bayes' theorem, the probability event independence, and the multiplication rule Conditional probability is a measure of the possibility of an event happening if another event has previously happened in probability theory. On the Vedantu website and the Vedantu app, you will find NCERT Solutions for Chapter 13 Probability of Class 12 Maths free of cost in a PDF format and you can download them for your exam purpose.

**6. What are the important topics covered in NCERT Solutions for Chapter 13 Probability of Class 12 Maths?**

All of the important characteristics and theorems of conditional probability are briefly described in Chapter 13 Probability of Class 12 Maths. Theorems and terms related to conditional probability, such as the multiplication theorem on probability, independent events, total probability, Bayes' theorem, random variables, and their probability distribution, mean and variance of a random variable, are among the important topics covered in the NCERT Solutions for Chapter 13 Probability of Class 12 Maths. Students can improve their fundamental understanding of probability by attempting all of the sums offered in these solutions.

**7. How many questions are there in NCERT Solutions for Chapter 13 Probability of Class 12 Maths?**

There are 81 questions in five exercises in Chapter 13 Probability of Class 12 Maths that ensure an accurate and clear comprehension of the entire course. Students can effectively strategize their preparation by examining the number of problems presented in each area and quickly mastering them. The format of these solutions is rather intuitive in conveying a thorough understanding of all important concepts and fundamentals of conditional probability. All these questions are available on Vedantu.

**8. State the important concepts which are discussed in NCERT Solutions for Chapter 13 Probability of Class 12 Maths?**

Students will gain a fundamental grasp of the formula for conditional probability and the theorems based on it, such as the multiplication theorem, theorem of total probability, and Bayes' theorem, by studying NCERT Solutions for Chapter 13 Probability of Class 12 Maths. A random variable, its probability distribution, mean and variance of a random variable, Bernoulli Trials, and Binomial Distribution are some of the more essential concepts and formulas contained in these answers. With the use of examples and drawings, these topics are thoroughly explained

**9. Why should I practise NCERT Solutions for Chapter 13 Probability of Class 12 Maths?**

NCERT Solutions for Chapter 13 Probability of Class 12 Maths are carefully crafted to provide a thorough grasp of probability's practical applications. Practising the sample issues presented in these solutions is based on our observations of probability in the real world. Working with these real-life scenarios can aid students in honing their problem-solving skills, which will increase their mental ability to deal with the most difficult situations. Visit Vedantu for solved questions of probability.