Chapter 13: Probability

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

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

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

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

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

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

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

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

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

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