Let `H_(1),H_(2), . . . ,H_(n)` be mutually exclusive and exhaustive events with `P(H_(i)) gt 0,i=1,2, . . .n`. Let E be any other event with `0 lt P(E) lt 1`.
STATEMENT-1: `P(H_(i)|E) gtP(E|H_(1)).P(H_(i))` for `i=1,2, . . .,n`
because
STATEMENT-2: `sum_(i=1)^(n)P(H_(i))=1`

Options

A. (a)Statement -1 is true , Statement -2 is true, Statement -2 is a correct explanation for Statement -1
B. (b)Statement -1 is true , Statement -2 is true, Statement -2 is not a correct explanation for Statement -1
C. (c)Statement -1 is true , Statement -2 is false
D. (d)Statement-1 is false, Statement-2 is true

Answer

To solve the problem, we need to analyze the statements provided and verify their validity based on the principles of probability.

### Step-by-Step Solution:

1. **Understanding the Events**:
- We have mutually exclusive and exhaustive events \( H_1, H_2, \ldots, H_n \) such that \( P(H_i) > 0 \) for \( i = 1, 2, \ldots, n \).
- The event \( E \) is any event with \( 0 < P(E) < 1 \).

2. **Statement 1**:
- We need to prove or disprove that \( P(H_i | E) > P(E | H_i) \cdot P(H_i) \) for \( i = 1, 2, \ldots, n \).
- By using Bayes' theorem, we can express \( P(H_i | E) \) as:
\[ P(H_i | E) = \frac{P(H_i \cap E)}{P(E)} \]
- Therefore, the inequality can be rewritten as:
\[ \frac{P(H_i \cap E)}{P(E)} > P(E | H_i) \cdot P(H_i) \]

3. **Using Bayes' Theorem**:
- From Bayes' theorem, we know that:
\[ P(E | H_i) = \frac{P(H_i \cap E)}{P(H_i)} \]
- Substituting this into our inequality gives:
\[ \frac{P(H_i \cap E)}{P(E)} > \frac{P(H_i \cap E)}{P(H_i)} \cdot P(H_i) \]
- This simplifies to:
\[ \frac{P(H_i \cap E)}{P(E)} > P(H_i \cap E) \]

4. **Rearranging the Inequality**:
- Rearranging gives:
\[ P(H_i \cap E) > P(E) \cdot P(H_i \cap E) \]
- Since \( P(H_i \cap E) > 0 \), we can divide both sides by \( P(H_i \cap E) \) (which is valid as long as it is not zero):
\[ 1 > P(E) \]
- This is always true since \( 0 < P(E) < 1 \).

5. **Conclusion for Statement 1**:
- Thus, we conclude that Statement 1 is true.

6. **Statement 2**:
- The statement is \( \sum_{i=1}^{n} P(H_i) = 1 \).
- This is a fundamental property of mutually exclusive and exhaustive events, which is always true.

7. **Final Conclusion**:
- Both statements are true, but Statement 2 does not serve as an explanation for Statement 1. Therefore, the correct conclusion is that both statements are true, but Statement 2 is not the correct explanation for Statement 1.

### Final Answer:
- Statement 1 is true.
- Statement 2 is true.
- Statement 2 is not a correct explanation for Statement 1.