NCERT Solutions for Class 12 Maths Chapter 13 Probability Miscellaneous Exercise

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

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

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

$\Rightarrow P\left( B\left| A \right. \right)=\dfrac{P\left( A \right)}{P\left( A \right)}=1$

(ii) $A\bigcap B=\phi $ 

Ans: since it is given that $A\bigcap B=\phi $

Therefore $\text{             }P\left( A\bigcap B \right)=0$

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

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 the following notations and their probabilities given as shown

N: Both children are males

$P\left( N \right)=\dfrac{1}{4}$

K: at least one of the children is male

\[P\left( K \right)=\dfrac{3}{4}\]

$\because N\bigcap K=BB$

$\therefore P\left( N\bigcap K \right)=\dfrac{1}{4}$

Now probability that both of the children are males provided at least one of the child is male is given by

$P\left( N\left| K \right. \right)=\dfrac{P\left( N\bigcap K \right)}{P\left( K \right)}$ 

$P\left( N\left| K \right. \right)=\dfrac{\dfrac{1}{4}}{\dfrac{3}{4}}$

Hence the probability that both of the children are males provided at least one of the children is male is $\dfrac{1}{3}$

(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 the following notations and their probabilities given as shown

N: Both children are females

$P\left( N \right)=\dfrac{1}{4}$

K: the elder child is a female

\[P\left( K \right)=\dfrac{2}{4}\]

$\because N\bigcap K=GG$

$\therefore P\left( N\bigcap K \right)=\dfrac{1}{4}$

Now probability that both of the children are males provided at least one of the children is male is given by

$P\left( N\left| K \right. \right)=\dfrac{P\left( N\bigcap K \right)}{P\left( K \right)}$ 

$P\left( N\left| K \right. \right)=\dfrac{\dfrac{1}{4}}{\dfrac{2}{4}}$

Hence the probability that both of the children are males provided at least one of the children is male is $\dfrac{1}{2}$

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 $0.25$ per cent 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 $\dfrac{5}{5.25}$

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 the 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 a binomial distribution with

$n=10,p=\dfrac{9}{10},q=\dfrac{1}{10}$

The probability that at least $6$ people are right handed is given by

\[{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \dfrac{9}{10} \right)}}^{k}}{{\left( \dfrac{1}{10} \right)}^{10-k}}\]

Therefore probability that at most $6$ people are right handed is given by

\[1-{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \dfrac{9}{10} \right)}}^{k}}{{\left( \dfrac{1}{10} \right)}^{10-k}}\]

5. 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 $=\dfrac{2}{7}$

6. Suppose we have four boxes. A, B, C and D contain coloured marbles as given below.

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 the following notations

 R be the event of drawing a 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

The total number of marbles is $40$

The number of red marbles is $15$

$\therefore P\left( R \right)=\dfrac{15}{40}$

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

$P\left( R\left| E \right. \right)=\dfrac{P\left( R\bigcap E \right)}{P\left( E \right)}$

$\Rightarrow P\left( R\left| E \right. \right)=\dfrac{\dfrac{1}{40}}{\dfrac{15}{40}}$

Hence E be the event of selecting the box A is $\dfrac{1}{15}$

Similarly, the probability that red marble is picked from box B is given by

$P\left( R\left| F \right. \right)=\dfrac{\dfrac{6}{40}}{\dfrac{15}{40}}$

Hence the probability that red marble is picked from box B is $\dfrac{2}{5}$

Similarly, the probability that red marble is picked from box C is given by

$P\left( R\left| G \right. \right)=\dfrac{\dfrac{8}{40}}{\dfrac{15}{40}}$

Hence the probability that red marble is picked from box C is $\dfrac{8}{15}$

7. Assume that the chances of the patient having a heart attack are $40$%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by $30$% and a prescription of certain drugs reduces its chances 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 the following notations

E: The events when the person took yoga and meditation courses

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

F: The events when the person took drugs

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

G: the person suffered a 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 a heart attack despite having yoga and meditation courses is given by

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

$P\left( E\left| G \right. \right)=\frac{0.5\times 0.28}{P0.5\times 0.28+0.5\times 0.3}$

Hence the probability that the found person has a heart attack despite having yoga and meditation courses is $\frac{14}{29}$

8. 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 the total number of determinants of second order entries being $0's\text{ or 1 }\!\!'\!\!\text{ s}$ is ${{2}^{4}}$

The value of the 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 the probability that the value of the determinant is positive is $\dfrac{3}{16}$.

9. 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 the following notations

E: A fails

F: B fails

Given 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

$P\left( E\left| F \right. \right)=\dfrac{0.15}{0.3}$ 

Hence the probability that A fails given B has failed is $0.5$

(ii) P (A fails alone)

Ans: The 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 $0.05$

10. 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. Find the probability that the transferred ball is black.

Ans:  Let us have the following notations that 

E: the red ball is transferred

$P\left( E \right)=\dfrac{3}{7}$

F: the black ball is transferred

$P\left( F \right)=\dfrac{4}{7}$

G: the red ball is drawn

When a red ball is transferred

$P\left( G\left| E \right. \right)=\dfrac{5}{10}$

Similarly, When a black ball is transferred

$P\left( G\left| F \right. \right)=\dfrac{4}{10}$

$P\left( F\left| G \right. \right)=\dfrac{\dfrac{4}{7}\times \dfrac{4}{10}}{\dfrac{4}{7}\times \dfrac{4}{10}+\dfrac{3}{7}\times \dfrac{5}{10}}$

Hence probability that the transferred ball is black is $\dfrac{16}{31}$

Choose the correct answer in each of the following:

11. If A and B are two events such that $P\left( A \right) \ne 0$ and $P\left( {B|A} \right) = 1$, then

(A) $A \subset B$

(B) $B \subset A$

(C) \[B = \phi \]

(D) $A = \phi $

Ans: Given that, $P\left( A \right) \ne 0$ and $P\left( {B|A} \right) = 1$

That is $P\left( {B|A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$

$ \Rightarrow \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = 1$

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

$\therefore A \subset B$

Option (A) is correct

12. If $P\left( {A|B} \right) > P\left( A \right)$, then which of the following is correct:

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

(B) $P\left( {A \cap B} \right) < P\left( A \right).P\left( B \right)$

(C) $P\left( {B|A} \right) > P\left( B \right)$

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

Ans: Given that, $P\left( {A|B} \right) > P\left( A \right)$

That is $\dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} > P\left( A \right)$

$P\left( {A \cap B} \right) > P\left( A \right).P\left( B \right)$

$\dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} > P\left( B \right)$

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

Therefore, Option (C) is correct.

13. If A and B are any two events such that P(A)+P(B) - P(A and B) = P(A), then

(A) $P(B|A) = 1$

(B) $P(A|B) = 1$

(C) $P(B|A) = 0$

(D) $P(A|B) = 0$

Ans: Given that, A and B are any two events and P(A)+P(B) - P(A and B) = P(A)

$  P(A) + P(B) - P(A \cap B) = P(A) $

$  P(B) - P(A \cap B) = 0 $

$  P(A \cap B) = P(B) $

$  \therefore P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}} $

= 1 

Therefore, Option (B) is correct.

Conclusion

Class 12 Maths Chapter 13 Miscellaneous Exercise solutions are important for understanding various concepts thoroughly. It covers a wide range of issues that call for the use of several methods and formulas. It's crucial to concentrate on comprehending the fundamental ideas behind every question, as opposed to simply learning the answers by heart. You must understand the theory underlying each concepts, practise frequently, and refer to solved examples.

Class 12 Maths Chapter 13: Exercises Breakdown

CBSE Class 12 Maths Chapter 13 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.