## NCERT Solutions For Probability Class 12 Chapter 13 - Free PDF Download

The NCERT Solutions for Maths Probability Class 12 Chapter 13 provides a comprehensive guide to understanding the concepts of probability, which is a fundamental aspect of statistics and mathematics. This chapter covers topics such as conditional probability, the multiplication rule, Bayes' theorem, random variables, and probability distributions including binomial and Poisson distributions. These solutions help students grasp theoretical concepts and apply them to solve various problems, enhancing their analytical and problem-solving skills crucial for board exams and competitive exams. The chapter ensures students build a strong foundation in probability, essential for higher education and real-world applications.

- 4.1Chapter 13 of NCERT Class 12 Mathematics deals with Probability, which is an essential concept in statistics and mathematics. Probability is a branch of mathematics that deals with the study of random events and their outcomes.The chapter includes various exercises to help students understand the concept of probability better. The exercises are as follows:Exercise 13.1: In this exercise, students are introduced to the basic concepts of probability, such as sample space, event, and probability.Exercise 13.2: In this exercise, students learn about the classical definition of probability and solve problems related to it.Exercise 13.3: In this exercise, students learn about the axiomatic approach to probability and the properties of probability.Miscellaneous Exercise: In this exercise, students solve additional problems related to probability, such as problems on conditional probability, independent events, and random variables.Overall, this chapter teaches students how to analyze random events and their outcomes using probability theory. By the end of this chapter, students should be able to calculate the probability of events, use the multiplication theorem of probability, and apply Bayes' theorem to solve real-world problems.

- 5.1Exercise 13.1

- 5.2Exercise 13.2

- 5.3Exercise 13.3

- 5.4Miscellaneous Exercise

- 6.113.1 Introduction

- 6.213.2 Conditional Probability

- 6.3Properties of Probability Distribution

## Glance on Class 12 Probability NCERT Solutions Chapter 13

Learn how the probability of one event (A) can be affected by the occurrence of another event (B). This is formalized using conditional probability and formulas to calculate it.

Multiplication Theorem on Probability helps find the probability of the intersection of two independent events (A and B) occurring together.

Events are considered independent if the occurrence of one doesn't influence the probability of the other. This chapter explores how to identify independent events.

Bayes' Theorem is a powerful tool for calculating conditional probability when you have additional information about the events involved.

Important Formulas:

Probability of an Event (A): P(A)

Conditional Probability of Event A given Event B: P(A|B)

Multiplication Theorem: P(A n B) = P(A) * P(B|A) (for independent events)

Bayes' Theorem: P(B|A) = [P(A|B) * P(B)] / P(A)

There are links to video tutorials explaining Chapter 13 Class 12 Probability for better understanding.

There are four fully solved exercises questions in Class 12 Maths, Chapter 13

## Access Exercise wise NCERT Solutions for Chapter 13 Maths Class 12

S.No. | Current Syllabus Exercises of Class 12 Maths Chapter 13 |

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4 | NCERT Solutions of Class 12 Maths Probability Miscellaneous Exercise |

## Exercises under NCERT Class 12 Maths Chapter 13 Probability

Chapter 13 of NCERT Class 12 Mathematics deals with Probability, which is an essential concept in statistics and mathematics. Probability is a branch of mathematics that deals with the study of random events and their outcomes.

The chapter includes various exercises to help students understand the concept of probability better. The exercises are as follows:

Exercise 13.1: In this exercise, students are introduced to the basic concepts of probability, such as sample space, event, and probability.

Exercise 13.2: In this exercise, students learn about the classical definition of probability and solve problems related to it.

Exercise 13.3: In this exercise, students learn about the axiomatic approach to probability and the properties of probability.

Miscellaneous Exercise: In this exercise, students solve additional problems related to probability, such as problems on conditional probability, independent events, and random variables.

Overall, this chapter teaches students how to analyze random events and their outcomes using probability theory. By the end of this chapter, students should be able to calculate the probability of events, use the multiplication theorem of probability, and apply Bayes' theorem to solve real-world problems.

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

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

### Miscellaneous Exercise

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

$P\left( A\left| B \right. \right)=1$

## Practice and Master the Concepts of The Chapter Probability with NCERT Solutions for Class 12 Maths

### 13.1 Introduction

The Class 12 Maths probability chapters cover the measurement of uncertainty in random events, the axiomatic approach by A. N. Kolmogorov, addition rule in probability, conditional probability, Bayes’ theorem, independence of events, multiplication rule in probability, probability distribution of random variables, mean and variance of probability distribution, and the binomial distribution as a discrete probability distribution.

### 13.2 Conditional Probability

In Probability Class 12 NCERT Solutions, you will learn about conditional probability, which is the probability of event B occurring given that event A has already taken place. If events A and B are independent, the probability of event B given event A is simply P(B). However, if events A and B are not independent, the probability of the intersection of A and B (the probability of both events occurring) is denoted by P(A and B).

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

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)

### 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

## Overview of Deleted Syllabus for CBSE Class 12 Maths Probability 2024-25

Chapter | Dropped Topics |

Probability | Exercise 13.6 - Random Variables and Its Probability Distributions |

Example - 22 and 23 | |

Exercise 13.6.1 - Probability Distribution of a Random Variable | |

Exercise 13.6.2 - Mean of Random Variables | |

Exercise 13.6.3 - Variance of a Random Variable | |

Exercise 13.7 - Bernoulli Trials and Binomial Distribution | |

Example - 34 and 35 | |

Miscellaneous Exercise - Question Numbers 5–7, 9–11 | |

Page Number 585-586 - Last 3 Points of the Summary |

## Access Exercises of Class 12 Maths Chapter 13 Probability

Exercise | Number of Questions |

Exercise 13.1 | 19 Questions and Solutions |

Exercise 13.2 | 19 Questions and Solutions |

Exercise 13.3 | 14 Questions and Solutions |

Miscellaneous Exercise | 13 Questions and Solutions |

## Conclusion

Finally, Vedantu's NCERT Solutions for Class 12 Maths Chapter Probability provide a thorough grasp of probability ideas. It is important to concentrate on comprehending the fundamental ideas, such as events, sample space, and probability rules. Exam papers from previous years usually have 3-5 questions on probability, therefore in order to do well on tests, you must fully understand the ideas and practise handling a variety of issues. These Vedantu-provided solutions can significantly help students become proficient in probability and do well in tests.

## Other Study Material for CBSE Class 12 Maths Chapter 13

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## NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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## Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

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## Important Related Links for NCERT Class 12 Maths

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## FAQs on NCERT Solutions for 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.

10. Is Class 12 Maths Probability difficult?

Probability builds on your prior mathematical knowledge while introducing new concepts. Effective conceptual knowledge can be attained by concentrated practice and a solid foundation in the subject matter.

Here are some pointers for thriving in probability in Class 12:

Make sure you comprehend fundamental ideas such as Venn diagrams, sets, and counting principles (combinations and permutations).

Before tackling more difficult probability questions, practice with simpler ones.

Don't just memorise the probability calculations; instead, concentrate on comprehending their underlying reasoning.

If you're having trouble, don't be hesitant to approach your teacher or tutor for assistance.

11. What is the hardest part of probability?

Although probability appears simple and is dependent on chance, there are a few ideas that can be difficult to understand:

Imagine flipping two coins as opposed to just one in a complex event. Calculating the likelihood of particular events—such as both landing on heads—can be trickier to do than with a single coin.

Conditional Probability: This is the study of how the probability of one event occurring (or not) influences the probability of another event. If you're not used to thinking in this way, it can get perplexing. For example, if you have previously drawn a red card (without replacing it), the likelihood of drawing another red card from the deck changes.

12. What is the formula for probability?

The most fundamental probability formula is P(A) = n(A) / n(S).

Each component means what follows:

P(A): This is the likelihood that event A will occur.

The number of outcomes that are advantageous to event A is indicated by the value n(A).

The total number of possible outcomes in the sample space is denoted by n(S).

To put it another way, the probability is determined by dividing the entire number of possible outcomes by the number of ways a successful outcome (event A) might occur.

13. What are the applications of probability?

NCERT Probability Class 12 is used in many different fields to help them make crucial decisions.

Probability is used in finance to manage risk, price financial instruments, and evaluate investment hazards. Insurance companies use probability to evaluate the risks involved in various policies and set insurance premiums.

Probability is used in science to test theories, analyse data, and simulate random events.

Another fundamental idea in machine learning algorithms is probability, which is applied to tasks like anomaly detection, prediction, and classification.

Probability is essential to artificial intelligence for planning, reasoning in complicated contexts, and making decisions under uncertainty.

Self-driving cars, for instance, utilise probability to decide how best to maneuver through traffic.

14. Who invented probability?

Probability theory began with Fermat and Pascal in the 17th century and was further upon by Huygens and Bernoulli in the 18th century. It wouldn't be quite correct to credit probability to a single inventor because it was a joint effort.