JEE Main 2026 — Jan 22, Shift 1 — Mathematics solution

JEE Main2026Jan 22, Shift 1Mathematics

If a random variable x has the probability distribution

$x 01234567 p (x) 02 k k 3 k 2 k^{2} 2 k k^{2} + k 7 k^{2}$

then $P (3 < x \leq 6)$ is equal to

Solution

$\sum p (x_{i}) = 1$ : $0 + 2 k + k + 3 k + 2 k^{2} + 2 k + (k^{2} + k) + 7 k^{2} = 9 k + 10 k^{2} = 1$ .

Solve $10 k^{2} + 9 k - 1 = 0$ : $k = \frac{- 9 + 81 + 40}{20} = \frac{- 9 + 11}{20} = \frac{1}{10}$ (taking the positive root).

$P (3 < x \leq 6) = p (4) + p (5) + p (6) = 2 k^{2} + 2 k + (k^{2} + k) = 3 k + 3 k^{2}$ .

With $k = 1/10$ : $3 (0.1) + 3 (0.01) = 0.30 + 0.03 = 0.33$ .

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