CBSE Class 12 Maths Chapter 13 Probability – NCERT Solutions Exercise 13.2 [2025]

Exercise 13.2

1. If $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{3}}{\mathbf{5}}$ and $\mathbf{P}\left( \mathbf{B} \right)=\dfrac{\mathbf{1}}{\mathbf{5}}$, find P $\mathbf{P}\left( \mathbf{A}\cap \mathbf{B} \right)$ if $\mathbf{A}$ and $\mathbf{B}$ are independent events.

Ans. We are provided that, 

$\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{3}}{\text{5}}$ and $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{5}}$.

Then we have

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$, since $\text{A}$ and $\text{B}$ are independent.

$\text{=}\dfrac{3}{5}\cdot \dfrac{1}{5}$

$\text{=}\dfrac{3}{25}$.

Hence, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{\text{3}}{\text{25}}$.

2. Two cards are drawn at random and without replacement from a pack of $\mathbf{52}$ playing cards. Find the probability that both the cards are black.

Ans. It is known that in the pack of $\text{52}$ cards, there are $\text{26}$ black cards.

Let $\text{P}\left( \text{A} \right)$ denote the probability that a black card is drawn in the first draw.

Therefore, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{26}}{\text{52}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Let suppose $\text{P}\left( \text{B} \right)$ denotes the probability that a black card is drawn in the second draw.

Therefore, $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{25}}{\text{51}}$.

Hence, the probability that both the cards drawn are black $\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{25}}{\text{51}}\text{=}\dfrac{\text{25}}{\text{102}}$.

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 $\mathbf{15}$ oranges out of which $\mathbf{12}$ are good and $\mathbf{3}$ 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

$\therefore P\left( B \right)=\frac{11}{14}$

C: the third orange is good

$\therefore P\left( C \right)=\frac{10}{13}$

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=$\frac{12}{15}\times \frac{11}{14}\times \frac{10}{13}$

Thus probability that the box is approved for sale=$\frac{44}{91}$=0.48

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.

$\mathbf{A}:$ head comes on the coin.

$\mathbf{B}:$ $\mathbf{3}$ appears in the die.

Ans. The sample space when a fair coin and an unbiased coin are tossed is

$\text{S=}\left\{ \left( \text{H,1} \right)\text{,}\left( \text{H,2} \right)\text{,}\left( \text{H,3} \right)\text{,}\left( \text{H,4} \right)\text{,}\left( \text{H,5} \right)\text{,}\left( \text{H,6} \right)\text{,}\left( \text{T,1} \right)\text{,}\left( \text{T,2} \right)\text{,}\left( \text{T,3} \right)\text{,}\left( \text{T,4} \right)\text{,}\left( \text{T,5} \right)\text{,}\left( \text{T,6} \right) \right\}$

Since, for the event $\text{A}$, head comes on the fair coin, so

$\text{A=}\left\{ \left( \text{H,1} \right)\text{,}\left( \text{H,2} \right)\text{,}\left( \text{H,3} \right)\text{,}\left( \text{H,4} \right)\text{,}\left( \text{H,5} \right)\text{,}\left( \text{H,6} \right) \right\}$

Therefore, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{6}}{\text{12}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Now, $\text{B}$ suggests the event of appearing $\text{3}$ in the die.

So, $\text{B=}\left\{ \left( \text{H,3} \right)\text{,}\left( \text{T,3} \right) \right\}$.

So, $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{2}}{\text{12}}\text{=}\dfrac{\text{1}}{\text{6}}$.

Thus, $\text{A}\cap \text{B=}\left\{ \left( \text{H,3} \right) \right\}$ and so,

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{1}{12}$.

Now, 

\[\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{12}}\text{=P}\left( \text{A}\cap \text{B} \right)\].

Hence, it is concluded that the events $\text{A}$ and $\text{B}$ are independent.

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

$\mathbf{A}:$ the number is even.

$\mathbf{B}:$ the number is red.

Ans. The sample space when a die is thrown is 

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

Therefore, when the number is even, then

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

$\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Also, when the number turns red, then

$\text{B=}\left\{ 1,2,3 \right\}$ and therefore,

$\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Thus, $\text{A}\cap \text{B=}\left\{ 2 \right\}$ and so,

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{\text{1}}{\text{6}}$.

$\Rightarrow \text{P}\left( \text{AB} \right)\text{=}\dfrac{\text{1}}{\text{6}}$.

Now, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{4}}\ne \dfrac{1}{6}$

That is, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \left( \text{AB} \right)$.

Hence, the events $\text{A}$ and $\text{B}$ are not independent.

6. Let $\mathbf{E}$ and $\mathbf{F}$ be events with $\mathbf{P}\left( \mathbf{E} \right)=\dfrac{\mathbf{3}}{\mathbf{5}},\mathbf{P}\left( \mathbf{F} \right)=\dfrac{\mathbf{3}}{\mathbf{10}}$ and $\mathbf{P}\left( \mathbf{E}\cap \mathbf{F} \right)=\dfrac{\mathbf{1}}{\mathbf{5}}$. Are $\mathbf{E}$ and $\mathbf{F}$ independent?

Ans. We have, $\text{P}\left( \text{E} \right)\text{=}\dfrac{\text{3}}{\text{5}}\text{,P}\left( \text{F} \right)\text{=}\dfrac{\text{3}}{\text{10}}$ and $\text{P}\left( \text{E}\cap \text{F} \right)\text{=}\dfrac{\text{1}}{\text{5}}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{10}}\text{=}\dfrac{\text{9}}{\text{50}}\ne \dfrac{\text{1}}{\text{5}}$.

Therefore, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\ne \text{P}\left( \text{E}\cap \text{F} \right)$.

That is, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\ne \text{P}\left( \text{EF} \right)$.

Hence, the events $\text{E}$ and $\text{F}$ are not independent.

7. Given that the events $\mathbf{A}$ and $\mathbf{B}$ are such that $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{1}}{\mathbf{2}},\mathbf{P}\left( \mathbf{A}\cup \mathbf{B} \right)=\dfrac{\mathbf{3}}{\mathbf{5}}$ and $\mathbf{P}\left( \mathbf{B} \right)=\mathbf{p}$. Find p if they are (i) mutually exclusive (ii) independent.

Ans.

(i) It is known that if two events $\text{A}$ and $\text{B}$ are mutually exclusive, then

$\text{A}\cap \text{B=}\phi $.

Therefore, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=0}$.

Now, we also know that, $\text{P}\left( \text{A}\cup \text{B} \right)\text{=P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

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

$\Rightarrow \text{p=}\dfrac{\text{3}}{\text{5}}\text{-}\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{10}}$.

(ii)  Since, the events $\text{A}$ and $\text{B}$ are independent, so 

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\text{p}$.

Now, we know that, $\text{P}\left( \text{A}\cup \text{B} \right)\text{=P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

$\Rightarrow \dfrac{\text{3}}{\text{5}}\text{=}\dfrac{\text{1}}{\text{2}}\text{+p-}\dfrac{\text{1}}{\text{2}}\text{p}$

$\Rightarrow \dfrac{\text{p}}{\text{2}}\text{=}\dfrac{\text{3}}{\text{5}}\text{-}\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{10}}$

$\Rightarrow \text{p=}\dfrac{\text{2}}{\text{10}}\text{=}\dfrac{\text{1}}{\text{5}}$.

8. Let $\mathbf{A}$ and $\mathbf{B}$ be independent events with $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{0}.\mathbf{3},\,\,\mathbf{P}\left( \mathbf{B} \right)=\mathbf{0}.\mathbf{4}$. Find 

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

Ans. We are provided that $\text{P}\left( \text{A} \right)\text{=0}\text{.3,}\,\,\text{P}\left( \text{B} \right)\text{=0}\text{.4}$.

Since the two events are independent, so 

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=0}\text{.3}\times \text{0}\text{.4=0}\text{.12}$.

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

Ans. We know that, $\text{P}\left( \text{A}\cup \text{B} \right)=\text{P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

Since, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=0}\text{.12}$, so

$\text{P}\left( \text{A}\cup \text{B} \right)=\text{0}\text{.3+0}\text{.4-0}\text{.12=0}\text{.58}$.

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

Ans. We know that, the conditional probability

$\text{P}\left( \text{A }\!\!|\!\!\text{ B} \right)\text{=}\dfrac{\text{P}\left( \text{A}\cap \text{B} \right)}{\text{P}\left( \text{B} \right)}$

$\Rightarrow \text{P}\left( \text{A }\!\!|\!\!\text{ B} \right)\text{=}\dfrac{0.12}{0.4}\text{=0}\text{.3}$.

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

Ans. We know that, the conditional probability

$\text{P}\left( \text{B }\!\!|\!\!\text{ A} \right)\text{=}\dfrac{\text{P}\left( \text{A}\cap \text{B} \right)}{\text{P}\left( \text{A} \right)}$

Therefore, $\text{P}\left( \text{B }\!\!|\!\!\text{ A} \right)\text{=}\dfrac{0.12}{0.3}\text{=0}\text{.4}$.

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

Ans. We are provided that, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{1}}{\text{4}}\text{,P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{2}}$ and $\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{\text{1}}{\text{8}}$.

Now, it is known that, $\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=P}{{\left( \text{A}\cup \text{B} \right)}^{\prime }}$.

Therefore,

$\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=1-P}\left( \text{A}\cup \text{B} \right)$.

Again, it is known that,

$\text{P}\left( \text{A}\cup \text{B} \right)=\text{P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$.

Thus,

\[\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=1-P}\left( \text{A}\cup \text{B} \right)\text{=1-}\left[ \text{P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right) \right]\]

$\text{=1-}\left[ \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{8} \right]$

$\text{=1-}\dfrac{5}{8}$

$\text{=}\dfrac{3}{8}$.

Hence, \[\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=}\dfrac{3}{8}\].

10. Events $\mathbf{A}$ and $\mathbf{B}$ are such that $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{1}}{\mathbf{2}},\mathbf{P}\left( \mathbf{B} \right)=\dfrac{\mathbf{7}}{\mathbf{12}}$, and $\mathbf{P}\left( \mathbf{{A}'}\cap \mathbf{{B}'} \right)=\dfrac{\mathbf{1}}{\mathbf{4}}$. State whether $\mathbf{A}$ and $\mathbf{B}$ are independent?

Ans. We are provided that, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{,P}\left( \text{B} \right)\text{=}\dfrac{\text{7}}{\text{12}}$, and $\text{P}\left( \text{{A}'}\cup \text{{B}'} \right)\text{=}\dfrac{\text{1}}{\text{4}}$.

It is known that, $\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=P}{{\left( \text{A}\cup \text{B} \right)}^{\prime }}$.

Therefore, 

$\text{P}\left( \text{{A}'}\cup \text{{B}'} \right)\text{=}\dfrac{1}{4}$

\[\text{P}{{\left( \text{A}\cap \text{B} \right)}^{\prime }}\text{=}\dfrac{1}{4}\]

$\Rightarrow 1-\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{1}{4}$

$\Rightarrow \text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{3}{4}$.

But, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\times \dfrac{7}{12}\text{=}\dfrac{7}{24}$.

Therefore, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \left( \text{A}\cap \text{B} \right)$.

Hence, the events $\text{A}$ and $\text{B}$ are not independent.

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find (i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B)

11. Given two independent events $\mathbf{A}$ and $\mathbf{B}$ such that $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{0}\mathbf{.3},\mathbf{P}\left( \mathbf{B} \right)=\mathbf{0}.\mathbf{6}$. Find

(i) $\mathbf{P}$($\mathbf{A}$ and $\mathbf{B}$)

Ans. Since, the events $\text{A}$and $\text{B}$ are independent, so

$\text{P}$($\mathbf{A}$ and $\text{B}$)$\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$

$\text{=0}\text{.3}\times \text{0}\text{.6=0}\text{.18}$.

(ii) $\mathbf{P}$($\mathbf{A}$ and not $\mathbf{B}$)

Ans. Note that, 

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

$\text{=0}\text{.3-0}\text{.18}$

$\text{=0}\text{.12}$.

(iii) $\mathbf{P}$($\mathbf{A}$ or $\mathbf{B}$)

Ans.

It is known that, 

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

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

$\text{=0}\text{.3+0}\text{.6-0}\text{.18}$

$\text{=0}\text{.72}$.

(iv) $\mathbf{P}$(neither $\mathbf{A}$ nor $\mathbf{B}$)

Ans. It is known that, 

$\text{P}$(neither $\text{A}$ nor $\text{B}$)$\text{=P}{{\left( \text{A}\cup \text{B} \right)}^{\prime }}$

$\text{=1-P}\left( \text{A}\cup \text{B} \right)$

$\text{=1-0}\text{.72}$

$\text{=0}\text{.28}$

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

Ans. The probability of having an odd number in tossing the die one times $\text{=}\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Again, the probability of having an even number $\text{=}\dfrac{3}{6}\text{=}\dfrac{1}{2}$.

Therefore, the probability of having an even number thrice $\text{=}\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\text{=}\dfrac{1}{8}$.

Thus, the probability of obtaining an od number at least one times

$\text{=1-}$probability of not having an odd number in any of the throws

$\text{=1-}$probability of having an even number three times

$\text{=1-}\dfrac{1}{8}$

$\text{=}\dfrac{7}{8}$.

13. Two balls are drawn at random with replacement from a box containing $\mathbf{10}$ black and $\mathbf{8}$ red balls. Find the probability that 

(i) both balls are red. 

(ii) first ball is black and second is red. 

(iii) one of them is black and other is red.

Ans.

(i) When both balls are red.

Ans. Note that, there is total $\text{18}$ balls, among them $\text{8}$ are red and $\text{10}$ are black.

Therefore, the probability of having a red ball on drawing the first time $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

The total count of balls remains the same as the balls are replaced.

So, the probability of having a red ball on drawing second times $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

Thus, the probability of having both the balls is red $\text{=}\dfrac{4}{9}\times \dfrac{4}{9}\text{=}\dfrac{16}{81}$.

(ii) When the first ball is black and the other is red.

Ans. The probability of having a black ball in first draw $\text{=}\dfrac{10}{18}\text{=}\dfrac{5}{9}$.

The total count of balls remains the same as the balls are replaced.

Then, the probability of having a red ball in the second draw $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

Thus, the probability of having the black and red balls in the first and second draw respectively 

$\text{=}\dfrac{5}{4}\times \dfrac{4}{9}\text{=}\dfrac{20}{81}$.

(iii) When one of the balls is black and the other is red.

Ans. The probability of having a red ball in first draw $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

The total count of balls remains the same as the balls are replaced.

Then, the probability of having a black ball in the second draw $\text{=}\dfrac{10}{18}\text{=}\dfrac{5}{9}$.

Thus, the probability of having the red and black balls in the first and second draw respectively 

$\text{=}\dfrac{4}{9}\times \dfrac{5}{4}\text{=}\dfrac{20}{81}$.

Therefore, the probability of having one ball as red and the other as black 

$\text{=}$ probability of obtaining the first ball is black and the other is red 

         $+$ probability of obtaining the first ball is red and the other is black

$\text{=}\dfrac{20}{81}+\dfrac{20}{81}$

$=\dfrac{40}{81}$.

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

(i) the problem is solved 

(ii) exactly one of them solves the problem.

Ans.

(i) the problem is solved.

Ans. Probability that $\text{A}$ can solve the problem, $\text{P}\left( \text{A} \right)\text{=}\dfrac{1}{2}$,

Probability that $\text{B}$ can solve the problem, $\text{P}\left( \text{B} \right)\text{=}\dfrac{1}{3}$.

It is given that the problem is solved by them independently.

Therefore, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\times \dfrac{1}{3}\text{=}\dfrac{1}{6}$.

Also, $\text{P}\left( {\text{{A}'}} \right)\text{=1-P}\left( \text{A} \right)\text{=1-}\dfrac{1}{2}\text{=}\dfrac{1}{2}$ and

$\text{P}\left( {\text{{B}'}} \right)\text{=1-P}\left( \text{B} \right)\text{=1-}\dfrac{1}{3}\text{=}\dfrac{2}{3}$.

Thus, the probability that $\text{A}$ or $\text{B}$ can solve the problem

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

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

$\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}$

$\text{=}\dfrac{4}{6}$

That is, the probability that the problem is solved $\text{=}\dfrac{2}{3}$.

(ii) exactly one of them solves that specific problem.

Ans. The probability that exactly one of $\text{A}$ and $\text{B}$ solves the problem, 

$\text{P}\left( \text{A} \right)\cdot \text{P}\left( {\text{{B}'}} \right)+\text{P}\left( \text{B} \right)\cdot \text{P}\left( {\text{{A}'}} \right)$

$\text{=}\dfrac{1}{2}\times \dfrac{2}{3}+\dfrac{1}{2}\times \dfrac{1}{3}$

$\text{=}\dfrac{1}{3}+\dfrac{1}{6}$

$\text{=}\dfrac{1}{2}$.

15. One card is drawn at random from a well-shuffled deck of $\mathbf{52}$ 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’

(ii) E : ‘the card drawn is black’ 

      F : ‘the card drawn is a king’ 

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

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

Ans:

(i) $\mathbf{E}:$ the card drawn is a spade.

     $\mathbf{F}:$ the card drawn is an ace.

Ans. It is known that there are $13$ spade and $4$ ace cards in a deck of $\text{52}$ cards.

Therefore, the probability that the card chosen is a spade,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{13}{52}\text{=}\dfrac{1}{4}$.

Also, the probability that the card chosen is an ace,

$\text{P}\left( \text{F} \right)\text{=}\dfrac{4}{52}\text{=}\dfrac{1}{13}$.

Since only one card is an ace as well as a spade, so

$\text{P}\left( \text{EF} \right)\text{=}\dfrac{1}{52}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{1}{4}\times \dfrac{1}{13}\text{=}\dfrac{1}{52}\text{=P}\left( \text{EF} \right)$.

Hence, $\text{E}$ and $\text{F}$ are independent events.

(ii) $\mathbf{E}:$ the card drawn is black.

      $\mathbf{F}:$ the card drawn is a king.

Ans. It is known that there are $26$ black and $4$ king cards in a deck of $\text{52}$ cards.

Therefore, the probability that the card chosen is black,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{26}{52}\text{=}\dfrac{1}{2}$.

Also, the probability that the card chosen is a king,

$\text{P}\left( \text{F} \right)\text{=}\dfrac{4}{52}\text{=}\dfrac{1}{13}$.

Since only two cards are black as well as king, so

$\text{P}\left( \text{EF} \right)\text{=}\dfrac{2}{52}\text{=}\dfrac{1}{26}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{1}{2}\times \dfrac{1}{13}\text{=}\dfrac{1}{26}\text{=P}\left( \text{EF} \right)$.

Hence, $\text{E}$ and $\text{F}$ are independent events

(iii) $\mathbf{E}:$ the card drawn is a king or queen.

       $\mathbf{F}:$ the card drawn is a queen or jack.

Ans. It is known that there are $4$ king, $4$ queen, and $4$ jack cards in a deck of $\text{52}$ cards.

Therefore, the probability that the card chosen is a king or queen,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{8}{52}\text{=}\dfrac{2}{13}$.

Also, the probability that the card chosen is a queen or jack,

$\text{P}\left( \text{F} \right)\text{=}\dfrac{8}{52}\text{=}\dfrac{2}{13}$.

Since only $4$ cards are a king or a queen, or queen or a jack, so

$\text{P}\left( \text{EF} \right)\text{=}\dfrac{4}{52}\text{=}\dfrac{1}{13}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{2}{13}\times \dfrac{2}{13}\text{=}\dfrac{4}{169}\ne \dfrac{1}{13}$.

Therefore, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\ne \text{P}\left( \text{EF} \right)$.

Hence, $\text{E}$ and $\text{F}$ are not independent events.

16. In a hostel, $\mathbf{60}%$of the students read Hindi newspaper, $\mathbf{40}%$ read English newspaper and $\mathbf{20}%$ read both Hindi and English newspapers. A student is selected at random. 

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

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper. 

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Ans:

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

Ans. Let $\text{H}$ and $\text{E}$ denote the students who read Hindi newspapers and the students who read English newspapers respectively.

It is provided that,

$\text{P}\left( \text{H} \right)\text{=60 }\!\!%\!\!\text{ =}\dfrac{60}{100}$$\text{=}\dfrac{3}{5}$,

$\text{P}\left( \text{E} \right)\text{=40 }\!\!%\!\!\text{ =}\dfrac{40}{100}\text{=}\dfrac{2}{5}$, and

$\text{P}\left( \text{H}\cap \text{E} \right)\text{=20 }\!\!%\!\!\text{ =}\dfrac{20}{100}\text{=}\dfrac{1}{5}$.

The probability that a student reads neither Hindi nor English newspaper is, $\text{P}{{\left( \text{H}\cup \text{E} \right)}^{\prime }}\text{=1-P}\left( \text{H}\cup \text{E} \right)$

$\text{=1-}\left\{ \text{P}\left( \text{H} \right)+\text{P}\left( \text{E} \right)-\text{P}\left( \text{H}\cap \text{E} \right) \right\}$

$\text{=1-}\left( \dfrac{3}{5}+\dfrac{2}{5}-\dfrac{1}{5} \right)$

$\text{=1-}\dfrac{4}{5}$

$\text{=}\dfrac{1}{5}$.

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper. 

Ans. The probability that a student who reads an English newspaper also reads Hindi newspapers,

$\text{P}\left( \text{E }\!\!|\!\!\text{ H} \right)\text{=}\dfrac{\text{P}\left( \text{E}\cap \text{H} \right)}{\text{P}\left( \text{H} \right)}$

$\text{=}\dfrac{\dfrac{1}{5}}{\dfrac{3}{5}}$

$\text{=}\dfrac{1}{3}$.

(c) If she reads English newspaper, find the probability that she reads Hindi newspapers.

Ans. The probability that a student reads Hindi newspaper also reads English newspapers,

$\text{P}\left( \text{H }\!\!|\!\!\text{ E} \right)\text{=}\dfrac{\text{P}\left( \text{H}\cap \text{E} \right)}{\text{P}\left( \text{E} \right)}$

$\text{=}\dfrac{\dfrac{1}{5}}{\dfrac{2}{5}}$

$=\dfrac{1}{2}$.

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

(A) 0 

(B) $\frac{1}{3}$

(C) $\frac{1}{12}$

(D) $\frac{1}{36}$

Ans. The sample space contains a total of $36$ outcomes when two dice are thrown.

$2$ is the only even number which is prime.

Therefore, if $\text{E}$ be the event of obtaining an even prime number on each 

die, then $\text{E=}\left\{ \left( 2,2 \right) \right\}$ and so,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{1}{36}$.

18. Two events $\mathbf{A}$ and $\mathbf{B}$ will be independent, if 

(A) $\mathbf{A}$ and $\mathbf{B}$ are mutually exclusive.

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

(C) $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{P}\left( \mathbf{B} \right)$

(D) $\mathbf{P}\left( \mathbf{A} \right)+\mathbf{P}\left( \mathbf{B} \right)=\mathbf{1}$

Ans. 

Correct Answer:

(b) It is known that two events $\text{A}$ and $\text{B}$ are independent, if \[\text{P}\left( \text{AB} \right)\text{=P}\left( \text{A} \right)\times \text{P}\left( \text{B} \right)\].

Now, $\text{P}\left( \text{{A}'{B}'} \right)\text{=}\left[ \text{1-P}\left( \text{A} \right) \right]\left[ \text{1-P}\left( \text{B} \right) \right]$$\Rightarrow \text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=1-P}\left( \text{A} \right)-\text{P}\left( \text{B} \right)+\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$

$\Rightarrow 1-\text{P}\left( \text{A}\cup \text{B} \right)\text{=1-P}\left( \text{A} \right)-\text{P}\left( \text{B} \right)+\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)$

\[\Rightarrow \text{P}\left( \text{A}\cup \text{B} \right)\text{=P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)\]

\[\Rightarrow \text{P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{AB} \right)\text{=P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)\]

$\Rightarrow \text{P}\left( \text{AB} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$.

Hence, this concludes that $\text{A}$ and $\text{B}$ are independent events.

Thus, option (b) is correct.

Incorrect Answers:

(a) 

Now, let $\text{P}\left( \text{A} \right)\text{=m}$ and $\text{P}\left( \text{B} \right)\text{=n}$, where $0<\text{m,}\,\text{n1}$.

If possible, let $\text{A}$ and $\text{B}$ are two mutually exclusive events.

Therefore, $\text{A}\cap \text{B=}\phi $.

$\Rightarrow \text{P}\left( \text{AB} \right)\text{=0}$.

But, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=mn}\ne \text{0}$.

Thus, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \text{P}\left( \text{AB} \right)$,

that is, $\text{A}$ and $\text{B}$ are not independent events.

Hence, option (a) is incorrect.

(c)

Let consider two events $\text{A}$ and $\text{B}$ such that

$\text{A=}\left\{ 1,3,5 \right\}$, $\text{B=}\left\{ 2,4,6 \right\}$ and the sample space $\text{S=}\left\{ 1,2,3,4,5,6 \right\}$.

Therefore, $\text{P}\left( \text{A} \right)\text{=}\dfrac{3}{6}\text{=}\dfrac{1}{2}$ and $\text{P}\left( \text{B} \right)\text{=}\dfrac{3}{6}\text{=}\dfrac{1}{2}$,

that is, $\text{P}\left( \text{A} \right)\text{=P}\left( \text{B} \right)$.

Also, $\text{A}\cap \text{B=}\left\{ \phi  \right\}$ and so, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=0}$.

That is, $\text{P}\left( \text{AB} \right)\text{=0}$.

Now, $\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\times \dfrac{1}{2}\text{=}\dfrac{1}{4}\ne 0$

Thus, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \text{P}\left( \text{AB} \right)$,

that is, $\text{A}$ and $\text{B}$ are not independent events.

Hence, option (c) is incorrect.

(d) 

Considering the previous example, it can be seen that,

\[\text{P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}+\dfrac{1}{2}\text{=1}\], but it has been proved that the events $\text{A}$ and $\text{B}$ are not independent.

Hence, the option (d) is incorrect.

Conclusion

This exercise provides a variety of problems that enhance your analytical and problem-solving skills. By understanding these concepts is not only essential for your exams but also for applying mathematical reasoning in real-life situations. By thoroughly practicing the problems in Class 12 13.2 Exercise, you are better prepared to tackle more advanced topics in probability and succeed in your academic journey.

Class 12 Maths Chapter 13: Exercises Breakdown

CBSE Class 12 Maths Chapter 13 Other Study Materials

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