Let \(H_{1}, H_{2}, \cdots, H_{n}\) be mutually exclusive and exhaustive events with \(P(H_{i})>0\), \(i=1,2, \ldots, n\). Let \(E\) be any other event with \(0 P(E \mid H_{i}) \cdot P(H_{i})\) for \(i=1,2, \ldots, n\). because STATEMENT-2 : \(\sum_{i=1}^{n} P(H_{i})=1\).

(D) Statement-1 is False, Statement-2 is True

(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

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Probability XII

Let

H_{1}, H_{2}, \dots, H_{n}

be mutually exclusive and exhaustive events with

P (H_{i}) > 0

i = 1, 2, \dots, n

. Let

E

be any other event with

0 < P (E) < 1

.

STATEMENT-1 :

P (H_{i} ∣ E) > P (E ∣ H_{i}) \cdot P (H_{i})

for

i = 1, 2, \dots, n

.

because

STATEMENT-2 :

\sum_{i = 1}^{n} P (H_{i}) = 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✓ Correct

Answer: (D)

Statement-2 is clearly true since

H_{1}, H_{2}, \dots, H_{n}

are exhaustive.

For Statement-1: By Bayes' theorem,

P (H_{i} ∣ E) = \frac{P ( E ∣ H _{i} ) P ( H _{i} )}{P ( E )}

Since

0 < P (E) < 1

, we have

P (H_{i} ∣ E) > P (E ∣ H_{i}) P (H_{i})

when

P (E) < 1

.

However, if

P (H_{i} \cap E) = 0

for some

i

, then both sides equal zero, so the inequality doesn't hold for all

i

.

Therefore, Statement-1 is False and Statement-2 is True.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Single correct option