A random variable X takes values 0,1,2,3 with probabilities frac2 a+130, frac8 a-130, frac4 a+130, b respectively, where a, b in mathbfR. 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 :

X 0 1 2 3 p(x) (2a + 1 / 30) (8a - 1 / 30) (4a + 1 / 30) b \ sigma^2=sum x_i^2 p(x_i)-mu^2 sigma^2+mu^2=sum x_i^2 p(x_i) =0+1((8 a-1 / 30))+4((4 a+1 / 30))+9 b => (24 a+270 b+3 / 30)=2 24 a+270 b=57 8 a+90 b=19 ... (1) Also \ sum p( i)=1 (2 a+1 / 30)+(8 a-1 / 30)+(4 a+1 / 30)+b=1 14 a+30 b=29 ......(2) Solving (1) (2) \ a=2, b=(1 / 30), frac a b=60

A random variable X takes values 0123 with probabilitie... | null (Mathematics)

JEE Mains 2026 21st January Shift 2

Mathematics

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 :

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Correct Answer:

Explanation

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

\\ \sigma^{2}=\sum x_{i}^{2} p\left(x_{i}\right)-\mu^{2} \\

\sigma^{2}+\mu^{2}=\sum x_{i}^{2} p\left(x_{i}\right) \\

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

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

24 a+270 b=57 \\

8 a+90 b=19 \ldots(1)

Also

\\ \sum \mathrm{p}(\mathrm{i})=1\\

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

14 \mathrm{a}+30 \mathrm{~b}=29 \tag......{(2)}\\

Solving (1) & (2)

\\ \mathrm{a}=2, \quad \mathrm{~b}=\frac{1}{30}, \frac{\mathrm{a}}{\mathrm{b}}=60

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