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

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Probability

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 :

Answer: (C)

We have

σ^{2} = \sum x_{i}^{2} p (x_{i}) - μ^{2}

, so

σ^{2} + μ^{2} = \sum x_{i}^{2} p (x_{i})

. Computing:

0^{2} \cdot \frac{2 a + 1}{30} + 1^{2} \cdot \frac{8 a - 1}{30} + 2^{2} \cdot \frac{4 a + 1}{30} + 3^{2} \cdot b = 2

. This gives

\frac{8 a - 1 + 16 a + 4}{30} + 9 b = 2

, or

\frac{24 a + 3}{30} + 9 b = 2

. Simplifying:

24 a + 270 b = 57

, or

8 a + 90 b = 19

(equation 1). Also,

\sum p (x_{i}) = 1

gives

\frac{2 a + 1 + 8 a - 1 + 4 a + 1}{30} + b = 1

, or

\frac{14 a + 1}{30} + b = 1

. Thus

14 a + 30 b = 29

(equation 2). Solving equations 1 and 2: from equation 2,

14 a = 29 - 30 b

. Substituting into equation 1:

8 a + 90 b = 19

becomes

\frac{8 ( 29 - 30 b )}{14} + 90 b = 19

, giving

\frac{232 - 240 b}{14} + 90 b = 19

. Multiplying by 14:

232 - 240 b + 1260 b = 266

, so

1020 b = 34

and

b = \frac{1}{30}

. Then

a = \frac{29 - 30 \cdot \frac{1}{30}}{14} = \frac{28}{14} = 2

. Thus

\frac{a}{b} = \frac{2}{1/30} = 60

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option