JEE Main 2026 Jan 21 Shift 2, Mathematics Q10: Introduction To Probability

JEE Main2026Jan 21, Shift 2Mathematics

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 :

A
30
B
3
C
60
D
12

Solution

Use $σ^{2} + μ^{2} = E [X^{2}] = \sum x_{i}^{2} p (x_{i})$ and the total probability constraint $\sum p (x_{i}) = 1$ .

$E [X^{2}] = 0^{2} \cdot p (0) + 1^{2} (\frac{8 a - 1}{30}) + 2^{2} (\frac{4 a + 1}{30}) + 3^{2} b = \frac{24 a + 3}{30} + 9 b = 2$ .

So $24 a + 270 b = 57$ , i.e. $8 a + 90 b = 19$ … (i)

$\sum p_{i} = 1$ : $\frac{( 2 a + 1 ) + ( 8 a - 1 ) + ( 4 a + 1 )}{30} + b = 1 \Rightarrow \frac{14 a + 1}{30} + b = 1$ , i.e. $14 a + 30 b = 29$ … (ii)

Solve (i)–(ii): from (i) $\times$ 1 and (ii) $\times$ 3, eliminate $b$ : $14 a + 30 b = 29$ and $8 a + 90 b = 19 \Rightarrow$ $42 a + 90 b = 87$ and $8 a + 90 b = 19$ , subtract: $34 a = 68 \Rightarrow a = 2$ . Then $b = \frac{1}{30}$ .

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

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JEE Main 2026 Jan 21 Shift 2, Mathematics Q10: Introduction To Probability

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