CBSE Class 12 Mathematics Solutions for Chapter 13 Probability – NCERT Exercise 13.1 [2025-26]

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  $\[P\left( E\cap F \right)=0.2\], 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 $P\left( E\cap F \right)=0.2$

Now P(E|F) is given by

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

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

Hence we found that 

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

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

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

Now we know that\[P\left( F\cap E \right)\] and $P\left( E\cap F \right)$ 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 $P\left( A\cap B \right)=0.32$

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

So to find P(A|B we use the formula 

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

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

$\frac{32}{50}$

$\frac{16}{25}$

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) $P\left( A\cap B \right)$

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

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

$\Rightarrow P\left( A\cap B \right)=P\left( A|B \right).P\left( B \right)$
$\Rightarrow P\left( A\cap B \right)=0.4\times 0.8$

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

(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

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

Now put $P\left( A\cap B \right)=0.32$, $P\left( B \right)=0.5$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

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

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

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

$\text{  }P\left( A\cap B \right)=0.32$ 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 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(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 

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

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

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

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

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

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

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

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

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

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

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

Therefore $P\left( E\cap F \right)=\frac{1}{8}$

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

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

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

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

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

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

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

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

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

Therefore $P\left( E\cap F \right)=\frac{3}{8}$

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

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

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

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

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

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

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

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

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

Therefore $P\left( E\cap F \right)=\frac{6}{8}$

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

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

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

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 $E\cap F$ is given by

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

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

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

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

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 $E\cap F$ is given by

$E\cap F=\phi $

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

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

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

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 appear respectively on the 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 $E\cap F=\left\{ \left. \left( 6,5,4 \right) \right\} \right.$

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

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

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

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

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

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

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

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

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

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

10. A black and a red dice are rolled.

1. 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 the black die and the second from the 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 the 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 $A\cap B=\left\{ \left( 5,5 \right),\left( 5,6 \right) \right\}$

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

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

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

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

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

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

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

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

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

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 $E\cap F=\left\{ 3 \right\}$

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

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

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

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

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

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

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 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Similarly 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We know that 

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

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

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

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 questions and 400 difficult multiple choice questions. 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

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

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

Probability for selecting multiple choice questions is given by

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

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

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

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

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

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

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

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

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

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\cap B=\phi $

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

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

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

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

1.  $0$

2.  $\frac{1}{2}$

3.  Not Defined

4.  $1$

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

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

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

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\cap B=\phi $) (D)= $P\left( A \right)=P\left( B \right)$

Ans: It is given in the question that  

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

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

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

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

Conclusion

NCERT Solutions for Maths Exercise 13.1 Class 12 Chapter 13 - Probability by Vedantu gives clear and simple explanations. Focus on learning the formulas for conditional probability and the multiplication rule. These are key for solving complex problems accurately. By practicing these exercises, students can improve their problem-solving skills and get a strong grasp of probability, which will help them do better in exams.

Class 12 Maths Chapter 13: Exercises Breakdown

CBSE Class 12 Maths Chapter 13 Other Study Materials

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