Mathematics · Probability

JEE Main 2025 — 3 April, Evening Shift — Question 27

If the probability that the random variable $X$ takes the value $x$ is given by

$P(X=x)=k(x+1) 3^{-x}, x=0,1,2,3 \ldots$ , where $k$ is a constant, then $P(X \geq 3)$ is equal to

Answer: C

$s=\frac{k}{3^{0}}+\frac{2 k}{3}+\frac{3 k}{3^{2}}+\ldots$

$\frac{s}{3}=\frac{k}{3}+\frac{2 k}{3^{2}}+\ldots$

$s-\frac{s}{3}=k+\frac{k}{3}+\frac{k}{3^{2}}+\ldots$

$\frac{2 s}{3}=k\left(1+\frac{1}{3}+\frac{1}{3^{2}}+\ldots\right)$

$\frac{2 s}{3}=k \times \frac{1}{1-\frac{1}{3}}=\frac{3 k}{2}$

$S=\frac{9 k}{4}=1 \quad$ (Total probability)

$k=\frac{4}{9}$

$P(x \geq 3)=1-(P(x=0)+P(x=1)+P(x=2))$

$=1-\left(k+\frac{2 k}{3}+\frac{3 k}{3^{2}}\right)$

$=1-2 \mathrm{k}$

$=1-2 \times \frac{4}{9}$

$=\frac{1}{9}$

Answer key and solution verified before publishing.

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