A random variable X takes values 0,1,2,3 with probabilities frac 2 a +1 30 , frac 8 a -1 30 , frac 4 a +1 30 , b respectively, where a , b in R . Let μ and σ respectively be the mean and standard deviation of X such that σ 2 +μ 2 =2 . Then frac a b is equal to :

Mathematics · Probability

JEE Main 2026 — 21 January, Evening Shift — Question 10

A random variable $X$ takes values $0,1,2,3$ with probabilities $\frac{2 \mathrm{a}+1}{30}, \frac{8 \mathrm{a}-1}{30}, \frac{4 \mathrm{a}+1}{30}$ , b respectively, where $\mathrm{a}, \mathrm{b} \in \mathbf{R}$ . Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of X such that $\sigma^{2}+\mu^{2}=2$ . Then $\frac{\mathrm{a}}{\mathrm{b}}$ is equal to :

Option A:
$30$
Option B:
$3$
Option C:
$60$
Correct
Option D:
$12$

Answer: C

Step-by-step solution

X	0	1	2	3
p(x)	$\dfrac{2a+1}{30}$	$\dfrac{8a-1}{30}$	$\dfrac{4a+1}{30}$	b

$\sigma^{2}=\sum \mathrm{x}_{\mathrm{i}}^{2} \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right)-\mu^{2}$ $\sigma^{2}+\mu^{2}=\sum \mathrm{x}_{\mathrm{i}}^{2} \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right)$ $=0+1\left(\frac{8 a-1}{30}\right)+4\left(\frac{4 a+1}{30}\right)+9 b$ $\Rightarrow \frac{24 \mathrm{a}+270 \mathrm{~b}+3}{30}=2$ $24 \mathrm{a}+270 \mathrm{~b}=57$ $\begin{gathered} 8 \mathrm{a}+90 \mathrm{~b}=19 \end{gathered}$ Also $\sum \mathrm{p}(\mathrm{i})=1$ $\frac{2 a+1}{30}+\frac{8 a-1}{30}+\frac{4 a+1}{30}+b=1$ $\begin{gathered} 14 \mathrm{a}+30 \mathrm{~b}=29 \end{gathered}$ Solving & (2) $\mathrm{a}=2, \quad \mathrm{~b}=\frac{1}{30}, \frac{\mathrm{a}}{\mathrm{b}}=60$

Answer key and solution verified before publishing.

Practise Probability

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: Probability
Topic: Mean, variance, expected values of distributions