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JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Probability

Let a random variable X take values

0, 1, 2, 3

with

P (X = 0) = P (X = 1) = p, P (X = 2) = P (X = 3)

and

E (X^{2}) = 2 E (X)

. Then the value of

8 p - 1

is :

(A)0
(B)2✓ Correct
(C)1
(D)3

Answer: (B)

Solution

Let

P (X = 0) = P (X = 1) = p

and

P (X = 2) = P (X = 3) = q

. Since the probabilities add to

1

,

2 p + 2 q = 1

, so

p + q = \frac{1}{2}

. Then

E (X) = p + 2 q + 3 q = p + 5 q

and

E (X^{2}) = p + 4 q + 9 q = p + 13 q

. The condition

E (X^{2}) = 2 E (X)

gives

p + 13 q = 2 p + 10 q

, so

p = 3 q

. With

p + q = \frac{1}{2}

this yields

q = \frac{1}{8}

and

p = \frac{3}{8}

. Therefore

8 p - 1 = 3 - 1 = 2

.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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