Chapter 13: Probability

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

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 \[P\left( A|B \right)=\frac{16}{25}\]

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 $P\left( A\bigcap B \right)=0.32$

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 $P\left( A|B \right)=0.64$

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 $P\left( A\bigcup B \right)=0.98$

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

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

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 \[P\left( A\left| B \right. \right)=\frac{4}{5}\]

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 $P\left( B\left| A \right. \right)=\frac{2}{3}$

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

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

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

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

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 $P\left( E\left| F \right. \right)=0$

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

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

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

1. 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 $P\left( E\left| F \right. \right)=\frac{1}{9}$