Question

$E_1$ and $E_2$ are two events such that $P\left(E_1\right)=\frac{1}{4}$, $P\left(\frac{E_2}{E_1}\right)=\frac{1}{2}$ and $P\left(\frac{E_1}{E_2}\right)=\frac{1}{4}$, then

• A. $E_1$ and $E_2$ are independent
• B. $E_1,E_2$ are exhaustive
• C. $P\left(E_2\right)=2P\left(E_1\right)$
• D. $P(E_1\cap E_2),P\left(E_2\right)$ and $P\left(E_1\right)$ are in G. P.

Answer 1

Answer: (A), (C)

$E_1$ and $E_2$ are independent
$P\left(E_2\right)=2P\left(E_1\right)$

Given :- $P\left(E_1\right)=\frac{1}{4},P\left(\frac{E_2}{E_1}\right)=\frac{1}{2}$ and $P\left(\frac{E_1}{E_2}\right)=\frac{1}{4}$
$P\left(\frac{E_2}{E_1}\right)=\frac{1}{2}=\frac{P(E_1\cap E_2)}{P\left(E_1\right)}\Rightarrow P(E_1\cap E_2)=\frac{1}{8}$
$P\left(\frac{E_1}{E_2}\right)=\frac{1}{4}=\frac{P(E_1\cap E_2)}{P\left(E_2\right)}\Rightarrow P\left(E_2\right)=\frac{1}{2}$
$\Rightarrow P\left(E_2\right)=2P\left(E_1\right)$
$\because P(E_1\cup E_2)=P\left(E_1\right)+P\left(E_2\right)-P(E_1\cap E_2)$ $=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\ne 1$
$\therefore E_1$ and $E_2$ are not exhaustive .
$P\left(E_1\right)\times P\left(E_2\right)=P(E_1\cap E_2)$. Hence, the events are independent.