Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P\left(E_1\right)=\frac{1}{4}$, $P\left(E_2 / E_1\right)=\frac{1}{2}$ and $P\left(E_1 / E_2\right)=\frac{1}{4}$, observe the lists given below
The correct matching of the List I from the List II is (A) (B) (C) (D)

Question

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P\left(E_1\right)=\frac{1}{4}$, $P\left(E_2 / E_1\right)=\frac{1}{2}$ and $P\left(E_1 / E_2\right)=\frac{1}{4}$, observe the lists given below

The correct matching of the List I from the List II is (A) (B) (C) (D)

Aicharya · Accepted Answer

(A) Given, $P\left(E_1ight)=\frac{1}{4}, P\left(\frac{E_1}{E_2}ight)=\frac{1}{4}$ and $P\left(\frac{E_2}{E_1}ight)=\frac{1}{2}$ $\begin{array}{ll}\Rightarrow & \frac{P\left(E_2 \cap E_1ight)}{P\left(E_1ight)}=\frac{1}{2} \ \Rightarrow & P\left(E_2 \cap E_1ight)=\frac{1}{8} \ 	ext { Also, } & P\left(\frac{E_1}{E_2}ight)=\frac{1}{4} \ \Rightarrow & \frac{P\left(E_1 \cap E_2ight)}{P\left(E_2ight)}=\frac{1}{4} \ \Rightarrow & \frac{1}{8 P\left(E_2ight)}=\frac{1}{4} \ \Rightarrow & P\left(E_2ight)=\frac{1}{2}\end{array}$ $\begin{aligned} (B) \quad P\left(E_1 \cup E_2ight) & =P\left(E_1ight)+P\left(E_2ight)-P\left(E_1 \cap E_2ight) \ & =\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{5}{8} \end{aligned}$ $\begin{aligned} (C) \quad P\left(\frac{\bar{E}_1}{\bar{E}_2}ight) & =\frac{P\left(\bar{E}_1 \cap \bar{E}_2ight)}{P\left(\bar{E}_2ight)} \ & =\frac{1-P\left(E_1 \cup E_2ight)}{1-P\left(E_2ight)}=\frac{1-\frac{5}{8}}{1-\frac{1}{2}} \ & =\frac{3}{4} \end{aligned}$ $\begin{aligned} (D) \quad P\left(\frac{E_1}{\bar{E}_2}ight) & =\frac{P\left(E_1 \cap \bar{E}_2ight)}{P\left(\bar{E}_2ight)} \ & =\frac{P\left(E_1ight)-P\left(E_1 \cap E_2ight)}{1-P\left(E_2ight)} \ & =\frac{\frac{1}{4}-\frac{1}{8}}{1-\frac{1}{2}}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4} \end{aligned}$

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P\left(E_1\right)=\frac{1}{4}$…

Solution