Question

Let $H_1,\dots ,H_n$ be mutually exclusive and exhaustive
events with $P\left(H_i\right)>0,i=1,2,\dots ,n.$ Let $E$ be any
other event with $0<P(E)<1.$
STATEMENT-1 : $P\left(H_i\mid E\right)>P\left(E\mid H_i\right)\cdot P\left(H_i\right)$ for $i=1$, $2,\dots ,n$
because
STATEMENT-2 : $\sum _{i=1}^{n}P\left(H_i\right)=1.$

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement- 1 is True, Statement- 2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

Answer: (D)

Statement-1: If $P\left(H_{i}nE\right)=0$ for some $i$, then
$P\left(\frac{H_i}{E}\right)=P\left(\frac{E}{H_i}\right)=0$
If $P\left(H_{i}nE\right)\ne 0$ for all $i=1,2,3,\dots ,n$, then
$P\left(\frac{H_i}{E}\right)=\frac{P\left(H_{i}nE\right)}{P\left(H_i\right)}\times \frac{P\left(H_i\right)}{P(E)}$
$=\frac{P\left(\frac{E}{H_i}\right)\times P\left(H_i\right)}{P(E)}$
$>P\left(\frac{E}{H_i}\right)\times P\left(H_i\right)$
Hence, Statement-$1$ may not always be true.
Statement-$2$: Clearly, we can write as
$K_1\cup H_2\cup H_3\cup \dots \cup H_n=S$
$\Rightarrow P\left(H_1\right)+P\left(H_2\right)+\dots +P\left(H_n\right)=1$
Hence, Statement-$2$ is true.