Question

Let $H_{1}, \ldots, H_{n}$ be mutually exclusive and exhaustive
events with $P\left(H_{i}\right)>0, i =1,2, \ldots, n .$ 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 $i =1$, $2, \ldots, n$
because
STATEMENT-2 : $\displaystyle\sum_{i=1}^{n} P\left(H_{i}\right)=1 .$

• 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 $P\left(H_{i} n E\right)=0$ for some $i$, then
$P\left(\frac{H_{i}}{E}\right)=P\left(\frac{E}{H_{i}}\right)=0$
If $P\left(H_{i} n E\right) \neq 0$ for all $i =1,2,3, \ldots, n$, then
$P\left(\frac{H_{i}}{E}\right)=\frac{P\left(H_{i} n E\right)}{P\left(H_{i}\right)} \times \frac{P\left(H_{i}\right)}{P(E)}$
$=\frac{P\left(\frac{E}{H_{i}}\right) \times P\left(H_{i}\right)}{P(E)}$
$ > 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
$K _{1} \cup H _{2} \cup H _{3} \cup \ldots \cup H _{n}= S$
$\Rightarrow P\left( H _{1}\right)+P\left( H _{2}\right)+\ldots+P\left( H _{n}\right)=1$
Hence, Statement-$2$ is true.