JEE Mathematics 2026 Question with Solution

Given: A random variable $X$ takes values $0,1,2,3$ with probabilities $\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b$ respectively.

Find: The value of $\frac{a}{b}$.

Using the relation

$$\sigma^2 = E(X^2) - [E(X)]^2$$

we get

$$\sigma^2 + \mu^2 = E(X^2)$$

Since it is given that $\sigma^2 + \mu^2 = 2$, therefore

$$E(X^2) = 2$$

Now, the sum of all probabilities must be $1$:

$$\frac{2a+1 + 8a-1 + 4a+1}{30} + b = 1$$ $$\frac{14a+1}{30} + b = 1$$ $$b = \frac{29-14a}{30}$$

Next,

$$E(X^2) = 0^2 \cdot P(0) + 1^2 \cdot P(1) + 2^2 \cdot P(2) + 3^2 \cdot P(3)$$ $$0 + 1 \cdot \frac{8a-1}{30} + 4 \cdot \frac{4a+1}{30} + 9b = 2$$ $$\frac{8a-1+16a+4}{30} + 9\left(\frac{29-14a}{30}\right) = 2$$ $$24a + 3 + 261 - 126a = 60$$ $$-102a = -204$$ $$a = 2$$

Now,

$$b = \frac{29-14(2)}{30} = \frac{1}{30}$$

Therefore,

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

So, the correct option is C.

Given: $\sigma^2 + \mu^2 = 2$.

Find: $\frac{a}{b}$.

The key shortcut is to use

$$\sigma^2 + \mu^2 = E(X^2)$$

directly, instead of calculating mean and variance separately.

From total probability,

$$\frac{14a+1}{30} + b = 1$$

so

$$b = \frac{29-14a}{30}$$

Now compute only the second moment:

$$1^2 \cdot \frac{8a-1}{30} + 2^2 \cdot \frac{4a+1}{30} + 3^2 b = 2$$ $$\frac{8a-1}{30} + \frac{16a+4}{30} + 9\left(\frac{29-14a}{30}\right) = 2$$ $$a = 2, \qquad b = \frac{1}{30}$$

Hence,

$$\frac{a}{b} = 60$$

Therefore, the correct option is C.