Analysis of Statement 2

1. Definition of Mutually Exclusive and Exhaustive Events:

   • The events $H_1, H_2, ..., H_n$ are given to be mutually exclusive, which means $H_i \cap H_j = \emptyset$ for any $i \neq j$.
   • They are also exhaustive, which means their union covers the entire sample space, i.e., $H_1 \cup H_2 \cup ... \cup H_n = S$.

2. Probability of the Sample Space:

   • By the axioms of probability, the probability of the entire sample space is 1, i.e., $P(S) = 1$.

3. Sum of Probabilities:

   • Since the events are exhaustive, $P(H_1 \cup H_2 \cup ... \cup H_n) = P(S) = 1$.
   • Since the events are mutually exclusive, the probability of their union is the sum of their individual probabilities: $P(H_1 \cup H_2 \cup ... \cup H_n) = \sum_{i=1}^{n} P(H_i)$
   • Combining these facts, we get: $\sum_{i=1}^{n} P(H_i) = 1$

4. Conclusion for Statement 2:

   • Statement 2 is a direct consequence of the definition of mutually exclusive and exhaustive events. Therefore, Statement 2 is True.

Analysis of Statement 1

1. The Inequality:

   • Statement 1 claims that $P(H_i | E) > P(E | H_i) \cdot P(H_i)$ for all $i=1, 2, ..., n$.

2. Applying Bayes' Theorem:

   • Bayes' theorem states that $P(H_i | E) = \frac{P(E | H_i) P(H_i)}{P(E)}$.
   • Substituting this into the inequality from Statement 1: $\frac{P(E | H_i) P(H_i)}{P(E)} > P(E | H_i) \cdot P(H_i)$

3. Analyzing the Inequality:

   • Let's analyze this inequality. We are given $P(H_i) > 0$.

   • Case 1: $P(E | H_i) > 0$ In this case, the term $P(E | H_i) P(H_i)$ is positive. We can divide both sides of the inequality by this term without changing the direction of the inequality: $\frac{1}{P(E)} > 1$ This is equivalent to $1 > P(E)$, or $P(E) < 1$. The problem states that $0 < P(E) < 1$, so this condition is met. Thus, if $P(E | H_i) > 0$, Statement 1 is true.

   • Case 2: $P(E | H_i) = 0$ This occurs if the event $E$ and the event $H_i$ are mutually exclusive, i.e., $E \cap H_i = \emptyset$. Since $E$ is described as 'any other event', this is a possibility. If $P(E | H_i) = 0$, the right side of the inequality in Statement 1 is $0 \cdot P(H_i) = 0$. The left side is $P(H_i | E) = \frac{P(H_i \cap E)}{P(E)} = \frac{P(E \cap H_i)}{P(E)} = \frac{0}{P(E)} = 0$ (since $P(E)>0$). The inequality becomes $0 > 0$, which is false.

4. Counterexample:

   • Since Statement 1 must hold for all $i=1, 2, ..., n$, we can show it is false by finding a single case where it fails.
   • Let the sample space be $S = \{1, 2, 3, 4\}$ with a uniform probability distribution, $P(\{k\}) = 1/4$ for each outcome $k$.
   • Let $H_1 = \{1, 2\}$ and $H_2 = \{3, 4\}$. These are mutually exclusive and exhaustive. $P(H_1) = 1/2 > 0$ and $P(H_2) = 1/2 > 0$.
   • Let the event $E = \{1, 2\}$. Then $P(E) = 1/2$, satisfying $0 < P(E) < 1$.
   • Now let's check the inequality for $i=2$.
     • $E \cap H_2 = \{1, 2\} \cap \{3, 4\} = \emptyset$. Therefore, $P(E \cap H_2) = 0$.
     • Right side of the inequality: $P(E | H_2) \cdot P(H_2) = \frac{P(E \cap H_2)}{P(H_2)} \cdot P(H_2) = P(E \cap H_2) = 0$.
     • Left side of the inequality: $P(H_2 | E) = \frac{P(H_2 \cap E)}{P(E)} = \frac{0}{1/2} = 0$.
     • The inequality for $i=2$ becomes $0 > 0$, which is false.

5. Conclusion for Statement 1:

   • Since we have found a valid scenario where the inequality does not hold for one of the events $H_i$, Statement 1 is False.

Final Conclusion

• Statement 1 is False.
• Statement 2 is True.

This corresponds to option D.