Q36 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
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 $\mathrm{a}, \mathrm{b} \in \mathbf{R}$. 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 :
- A$12$
- B$30$
- Correct answerC$60$
- D3
Solution
Sum of probabilities $= 1$: $\frac{2a+1+8a-1+4a+1}{30} + b = 1 \Rightarrow b = \frac{29-14a}{30}$.
$\sigma^2 + \mu^2 = E(X^2) = 2$.
$E(X^2) = \frac{8a-1}{30} + \frac{4(4a+1)}{30} + \frac{9(29-14a)}{30} = \frac{-102a + 264}{30} = 2$.
$-102a + 264 = 60 \Rightarrow a = 2$.
$b = \frac{29 - 28}{30} = \frac{1}{30}$.
$\frac{a}{b} = \frac{2}{1/30} = 60$.
$\sigma^2 + \mu^2 = E(X^2) = 2$.
$E(X^2) = \frac{8a-1}{30} + \frac{4(4a+1)}{30} + \frac{9(29-14a)}{30} = \frac{-102a + 264}{30} = 2$.
$-102a + 264 = 60 \Rightarrow a = 2$.
$b = \frac{29 - 28}{30} = \frac{1}{30}$.
$\frac{a}{b} = \frac{2}{1/30} = 60$.
Hence, the correct answer is option: 'C'.