JEE Main 2025 Apr 7 Shift 2, Mathematics Q6: Introduction To Probability

JEE Main2025Apr 7, Shift 2Mathematics

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 :

Solution

Let $P (X = 2) = P (X = 3) = q$ . Since all probabilities sum to 1: $2 p + 2 q = 1$ , i.e. $p + q = \frac{1}{2}$ .

$E (X^{2}) = 0^{2} p + 1^{2} p + 2^{2} q + 3^{2} q = p + 13 q$ , and $E (X) = 0 \cdot p + 1 \cdot p + 2 q + 3 q = p + 5 q$ .

The condition $E (X^{2}) = 2 E (X)$ gives $p + 13 q = 2 (p + 5 q) \Rightarrow 3 q = p$ .

Substituting into $p + q = \frac{1}{2}$ : $3 q + q = \frac{1}{2} \Rightarrow q = \frac{1}{8}, p = \frac{3}{8}$ .

$∴ 8 p - 1 = 8 (\frac{3}{8}) - 1 = 3 - 1 = 2$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q6: Introduction To Probability

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