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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Probability

If a random variable

x

has the probability distribution

x	0	1	2	3	4	5	6	7
$p (x)$	0	$2 k$	$k$	$3 k$	$2 k^{2}$	$2 k$	$k^{2} + k$	$7 k^{2}$

then

P (3 < x \leq 6)

is equal to

(A)0.34
(B)0.22
(C)0.64
(D)0.33✓ Correct

Answer: (D)

Solution

Normalising,

\sum p (x) = 10 k^{2} + 9 k = 1

, so

10 k^{2} + 9 k - 1 = 0

, whose admissible (positive) root is

k = \frac{1}{10}

. Then

P (3 < x \leq 6) = p (4) + p (5) + p (6) = 2 k^{2} + 2 k + (k^{2} + k) = 3 k^{2} + 3 k = \frac{3}{100} + \frac{3}{10} = \frac{33}{100} = 0.33.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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