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A random variable X takes values 0,1,2,3 with probabilities

\[ \frac{2 \mathrm{a}+1}{30}, \frac{8 \mathrm{a}-1}{30}, \frac{4 \mathrm{a}+1}{30}, \mathrm{~b} \text { respectively, where } \mathrm{a}, \mathrm{~b} \in \mathrm{R} . \]

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 [JEE MAIN 21 Jan 2026 S II]
(1) 30
(2) 3
(3) 60
(4) 12

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Solution

[t]{|l|l|l|l|l|} x	0	1	2	3
P(x)	$\frac{2 \mathrm{a}+1}{30}$	$\frac{8 \mathrm{a}-1}{30}$	$\frac{4 \mathrm{a}+1}{30}$	b

\[ \begin{aligned} & \sum P(x)=\frac{2 a+1}{30}+\frac{8 a-1}{30}+\frac{4 a+1}{30}+b=1 \\ & \Rightarrow 14 a+1+30 b=30 \Rightarrow 14 a+30 b=29 \ldots \\ & \text { Variance }=\sigma^{2}=\frac{\sum x_{i}^{2} P\left(x_{i}\right)}{\sum P\left(x_{i}\right)}-\mu^{2} \\ & \Rightarrow \sigma^{2}+\mu^{2}=\frac{\sum x_{i}^{2}\left(P\left(x_{i}\right)\right)}{1} \\ & \Rightarrow 2=0+1\left(\frac{8 a-1}{30}\right)+4\left(\frac{4 a+1}{30}\right)+9 b \\ & \Rightarrow 2=\frac{8 a-1+16 a+4+270 b}{30} \\ & \Rightarrow 60=24 a+270 b+3 \Rightarrow 24 a+270 b=57 \\ & \Rightarrow 8 a+90 b=19 \ldots(2) \\ & (\text { Eq. }(1)) 3-\text { Eq. }(2) \\ & 42 a+90 b-8 a-90 b=29 \times 3-19 \\ & \Rightarrow 34 a=87-19=68 \Rightarrow a=2, b=\frac{1}{30} \\ & \frac{a}{b}=\frac{2}{1 / 30}=60 \text { Answer }(3) \end{aligned} \]

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