Conditional Probability of Exhaustive Events and Given Event

To evaluate the given statements:

Evaluation of STATEMENT-2:
Since $H_1, H_2, \dots, H_n$ are given to be mutually exclusive and exhaustive events, their union is the entire sample space $S$, and they are pairwise disjoint. Therefore:
$P\left(\bigcup_{i=1}^n H_i\right) = \sum_{i=1}^n P(H_i) = P(S) = 1$
Hence, STATEMENT-2 is True.

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Evaluation of STATEMENT-1:
By the definition of conditional probability:
$P(H_i \mid E) = \frac{P(H_i \cap E)}{P(E)} = \frac{P(E \mid H_i) \cdot P(H_i)}{P(E)}$

We are given that $0 < P(E) < 1$, which implies:
$\frac{1}{P(E)} > 1$

However, $E$ is any event with $0 < P(E) < 1$. It is entirely possible that for some $i$, the event $E$ and $H_i$ are mutually disjoint, i.e., $E \cap H_i = \emptyset$. In that case:
$P(E \cap H_i) = 0 \implies P(E \mid H_i) = 0 \quad \text{and} \quad P(H_i \mid E) = 0$

Substituting these values into the inequality:
$P(H_i \mid E) > P(E \mid H_i) \cdot P(H_i) \implies 0 > 0$
which is mathematically false.

Therefore, the inequality does not necessarily hold for all $i = 1, 2, \dots, n$.
Hence, STATEMENT-1 is False.

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Conclusion:
Statement-1 is False, and Statement-2 is True.

Correct Option: (D)