JEE Advanced 2024 — Paper 1 — MATH-III solution

JEE Advanced2024Paper 1MATH-III

Let $X$ be a random variable, and let $P (X = x)$ denote the probability that $X$ takes the value $x$ . Suppose that the points $(x, P (X = x))$ , $x = 0, 1, 2, 3, 4$ , lie on a fixed straight line in the $x y$ -plane, and $P (X = x) = 0$ for all $x \in R - {0, 1, 2, 3, 4}$ . If the mean of $X$ is $\frac{5}{2}$ , and the variance of $X$ is $α$ , then the value of $24 α$ is ______.

Your answer

Solution

Since the points are collinear, $P (x) = P (0) + K x$ for $x = 0, 1, 2, 3, 4$ . Two constraints:

Total probability: $\sum_{x = 0}^{4} P (x) = 5 P (0) + 10 K = 1$ .

Mean: $\sum_{x = 0}^{4} x P (x) = 10 P (0) + 30 K = \frac{5}{2}$ .

Subtracting twice the first from the second: $10 K = \frac{1}{2}$ , giving $K = \frac{1}{20}$ and $P (0) = \frac{1}{10}$ .

So $P (0) = \frac{2}{20}$ , $P (1) = \frac{3}{20}$ , $P (2) = \frac{4}{20}$ , $P (3) = \frac{5}{20}$ , $P (4) = \frac{6}{20}$ .

$E [X^{2}] = \frac{1}{20} (0 + 3 + 16 + 45 + 96) = \frac{160}{20} = 8$

Variance $α = E [X^{2}] - (E [X])^{2} = 8 - \frac{25}{4} = \frac{7}{4}$ .

$∴ 24 α = 24 \times \frac{7}{4} = 42$

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