JEE Advanced 2024 Paper 1 · Q13 · Probability Distributions

Question

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 $xy$-plane, and $P(X = x) = 0$ for all $x \in \mathbb{R} - \{0, 1, 2, 3, 4\}$. If the mean of $X$ is $\dfrac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24\alpha$ is ______.

SolveKar AI · Accepted Answer

Let $P(X=x)=mx+c$, $x=0,1,2,3,4$. $\sum P=1\Rightarrow \sum_{x=0}^{4}(mx+c)=10m+5c=1$. $E[X]=\sum x(mx+c)=m\sum x^2+c\sum x=30m+10c=\dfrac{5}{2}$. Solve: from $10m+5c=1$ get $c=\frac{1-10m}{5}$. Substitute: $30m+10\cdot\frac{1-10m}{5}=\frac{5}{2}\Rightarrow 30m+2-20m=\frac{5}{2}\Rightarrow 10m=\frac{1}{2}\Rightarrow m=\frac{1}{20}$, $c=\frac{1-1/2}{5}=\frac{1}{10}$. Then $E[X^2]=\sum x^2 P=m\sum x^3+c\sum x^2=100m+30c=5+3=... $ compute: $\sum x^3=0+1+8+27+64=100$, $\sum x^2=30$. $E[X^2]=100\cdot\frac{1}{20}+30\cdot\frac{1}{10}=5+3=8$. Wait, but we need $\alpha=\mathrm{Var}(X)=E[X^2]-(E[X])^2=8-\frac{25}{4}=\frac{7}{4}$. Then $24\alpha=24\cdot\frac{7}{4}=42$.

JEE Advanced 2024 Paper 1 · Q13 · Probability Distributions

Solution