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 $\[P\left( E\bigcap F \right)=0.2\], find P(E|F) and P(F|E).
   (2023) Ans
   Solution:

   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 $P\left( F|E \right)=\frac{1}{3}$
2. Compute P(A|B), if $P\left( B \right)=0.5$ and $P\left( A\bigcap B \right)=0.32$
   (2021C) Ans
   Solution:

   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 \[P\left( A|B \right)=\frac{16}{25}\]
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) $P\left( A\bigcap B \right)$
   Ans
   Solution:

   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 $P\left( A\bigcap B \right)=0.32$ (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 $\therefore P\left( A|B \right)=\frac{P\left( A\bigcap B \right)}{P\left( B \right)}$ 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 $P\left( A|B \right)=0.64$ (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 $\text{ }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$ 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 $P\left( A\bigcup B \right)=0.98$