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 : [2026]

Question

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 : [2026]

Smart Achievers · Accepted Answer

(3)

$x$	0	1	2	3
$p (x)$	$\frac{2 a + 1}{30}$	$a, b \in R$	$\frac{4 a + 1}{30}$	$b$

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

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

$= 0 + 1 (\frac{8 a - 1}{30}) + 4 (\frac{4 a + 1}{30}) + 9 b$

$\Rightarrow \frac{24 a + 270 b + 3}{30} = 2$

$24 a + 270 b = 57$

$8 a + 90 b = 19 . . . (1)$

Also,

$\sum p (i) = 1$

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

$14 a + 30 b = 29 . . . (2)$

Solving (1) and (2),

$a = 2, b = \frac{1}{30}, \frac{a}{b} = 60$

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