Let $H_1, H_2, \ldots, H_n$ be mutually exclusive and exhaustive events with $P(H_i) > 0$, $i = 1, 2, \ldots, n$. Let E be any other event with $0 < P(E) < 1$. STATEMENT I: $P(H_i|E) > P(E|H_i) \cdot P(H_i)$ for $i = 1, 2, \ldots, n$ because STATEMENT II: $\sum_{i=1}^{n} P(H_i) = 1$.

Let H₁, H₂, …, Hₙ be mutually exclusive and exhaustive events with P(Hᵢ) > 0, i = 1, 2, …, n. Let E be any other event with 0 < P(E) < 1. STATEMENT I: P(Hᵢ|E) > P(E|Hᵢ) · P(Hᵢ) for i = 1, 2, …, n because STATEMENT II: ∑ᵢ₌₁ⁿ P(Hᵢ) = 1.

$H_1,\ldots,H_n \text{ mutually exclusive and exhaustive},\quad P(H_i) > 0$

$0 < P(E) < 1$

$\sum_{i=1}^{n} P(H_i) = 1$

$\text{Statement I: } P(H_i|E) > P(E|H_i)\,P(H_i)\ \text{for all } i$

$\text{Statement II: } \sum_{i=1}^{n} P(H_i) = 1$

$\text{To check: truth of I, II, and whether II explains I}$

$1.\ P(H_i|E) = \dfrac{P(E|H_i)P(H_i)}{P(E)}$ $2.\ P(E|H_i)P(H_i) = P(E \cap H_i)$ $3.\ 0 < P(E) < 1 \Rightarrow \text{compare};\ \text{role of } \sum P(H_i)=1$ $P(H_i|E) = \dfrac{P(E|H_i)\,P(H_i)}{P(E)}$ $P(E|H_i)\,P(H_i) = P(E \cap H_i)$ $P(H_i|E) = \dfrac{P(E \cap H_i)}{P(E)}$ $0 < P(E) < 1 \Rightarrow \tfrac{1}{P(E)} > 1$ $P(H_i|E) = \dfrac{P(E \cap H_i)}{P(E)} > P(E \cap H_i) = P(E|H_i)P(H_i)$ $\Rightarrow \text{Statement I is TRUE (strict, since } P(E) < 1\text{)}$ $\text{Exhaustive} + \text{mutually exclusive} \Rightarrow \sum_{i=1}^{n} P(H_i) = 1$ $\Rightarrow \text{Statement II is TRUE}$ $\text{I's reason: } P(E) < 1 \Rightarrow \tfrac{1}{P(E)} > 1$ $\sum P(H_i)=1 \ \not\Rightarrow\ P(E) < 1$ $P(E) = \sum_{i=1}^{n} P(E|H_i)\,P(H_i)\quad (\text{this uses II})$ $\text{but the inequality needs } P(E) < 1,\ \text{not the sum}$ $P(E)=1 \Rightarrow P(H_i|E) = P(E|H_i)P(H_i)\ \text{(equality)}$ $P(E)<1 \Rightarrow P(H_i|E) > P(E|H_i)P(H_i)\ \text{(strict)}$ $\text{I: True},\quad \text{II: True},\quad \text{II is NOT the explanation for I}$ $\boxed{\text{Answer: (2)}}$