JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Probability

Let

A, B

and C be three events such that the probability that exactly one of A and B occurs is

(1 - k)

, the probability that exactly one of

B

and

C

occurs is

(1 - 2 k)

, the probability that exactly one of C and A occurs is

(1 - k)

and the probability of all

A, B

and C occur simultaneously is

k^{2}

, where

0 < k < 1

. Then the probability that at least one of A ,

B

and

C

occur is :

Answer: (A)

Solution

Let

P (A) = a

P (B) = b

P (C) = c

P (A \cap B) = p

P (B \cap C) = q

P (C \cap A) = r

, and

P (A \cap B \cap C) = k^{2}

.

Exactly one of A and B:

P (A △ B) = a + b - 2 p = 1 - k

Exactly one of B and C:

P (B △ C) = b + c - 2 q = 1 - 2 k

Exactly one of C and A:

P (C △ A) = c + a - 2 r = 1 - k

Using inclusion-exclusion:

P (A \cup B \cup C) = a + b + c - p - q - r + k^{2}

From the three equations:

2 (a + b + c) - 2 (p + q + r) = 3 - 4 k

a + b + c - (p + q + r) = \frac{3 - 4 k}{2}

Therefore:

P (A \cup B \cup C) = \frac{3 - 4 k}{2} + k^{2}

For

0 < k < 1

, let

f (k) = \frac{3 - 4 k}{2} + k^{2} = k^{2} - 2 k + \frac{3}{2}

f^{'} (k) = 2 k - 2 < 0

for

k < 1

, so

f

is decreasing.

f (0) = \frac{3}{2} > \frac{1}{2}

and

f (1) = 1 - 2 + \frac{3}{2} = \frac{1}{2}

.

Since

0 < k < 1

, we have

P (A \cup B \cup C) > \frac{1}{2}

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option