Question

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

• A. $E_1$ and $E_2$ are independent
• B. $E_1$ and $E_2$ are exhaustive
• C. $E_2$ is twice as likely to occur as $E_1$
• D. probabilities of the events $E_1\cap E_2,E_1$ and $E_2$ are in G.P.

Answer 1

Answer: (A), (C), (D)

The correct options are
A $E_1$ and $E_2$ are independent
C $E_2$ is twice as likely to occur as $E_1$
D probabilities of the events $E_1\cap E_2,E_1$ and $E_2$ are in G.P.

Given $P\left(E_1\right)$$=1/4$, $P\left(E_2/E_1\right)$$=1/2$ and $P\left(E_1/E_2\right)$$=1/4$

$P\left(E_2/E_1\right)=P(E_2\cap E_1)/P\left(E_1\right)$

$\frac{1}{2}=\frac{P(E_2\cap E_1)}{\left(1/4\right)}$

$\frac{1}{8}=P(E_2\cap E_1)$

$P(E_2\cap E_1)=\frac{1}{8}$

Similarly, $P\left(E_1/E_2\right)=P(E_1\cap E_2)/P\left(E_2\right)$

$\frac{1}{4}=\frac{\left(1/8\right)}{P\left(E_2\right)}$

$P\left(E_2\right)=\frac{1}{2}$

Consider, $P\left(E_1\right).P\left(E_2\right)=\left(\frac{1}{4}\right)\ast \left(\frac{1}{2}\right)=\frac{1}{8}$

$P(E_2\cap E_1)=\frac{1}{8}$

$\Rightarrow P(E_1\cap E_2)=P\left(E_1\right).P\left(E_2\right)$

Therefore $E_1$ and $E_2$ are Independent events.

Since $P\left(E_2\right)=2\times P\left(E_1\right)$

$\Rightarrow$ $E_2$ is twice as likely to occur as $E_1$ in the above condition

Since, $\left(P\right(E_1){)}^2=P(E_2)\ast P(E_1\cap E_2)$

$\Rightarrow$ the events $P(E_1\cap E_2),P\left(E_1\right)$ and $P\left(E_2\right)$ are in G.P.