CUET UG 2025 — Mathematics Algebra

Given that 60% of committee members favor the proposal and 40% oppose it. A member is selected with random variable $X = 1$X=1 if favoring and $X = 0$X=0 if opposing.

The probabilities are:

$$P(X = 1) = 0.6 = \frac{3}{5}$$P(X=1)=0.6=53​

$$P(X = 0) = 0.4 = \frac{2}{5}$$P(X=0)=0.4=52​

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The expected value is:

$$E(X) = 0 \times P(X=0) + 1 \times P(X=1)$$E(X)=0×P(X=0)+1×P(X=1)

$$E(X) = 0 \times \frac{2}{5} + 1 \times \frac{3}{5}$$E(X)=0×52​+1×53​

$$E(X) = \frac{3}{5}$$E(X)=53​

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The expected value of $X^2$X2 is:

$$E(X^2) = 0^2 \times P(X=0) + 1^2 \times P(X=1)$$E(X2)=02×P(X=0)+12×P(X=1)

$$E(X^2) = 0 \times \frac{2}{5} + 1 \times \frac{3}{5}$$E(X2)=0×52​+1×53​

$$E(X^2) = \frac{3}{5}$$E(X2)=53​

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The variance is:

$$\text{Var}(X) = E(X^2) - [E(X)]^2$$Var(X)=E(X2)−[E(X)]2

$$\text{Var}(X) = \frac{3}{5} - \left(\frac{3}{5}\right)^2$$Var(X)=53​−(53​)2

$$\text{Var}(X) = \frac{3}{5} - \frac{9}{25}$$Var(X)=53​−259​

$$\text{Var}(X) = \frac{15}{25} - \frac{9}{25}$$Var(X)=2515​−259​

$$\text{Var}(X) = \frac{6}{25}$$Var(X)=256​

Therefore, the variance of the random variable $X$X is $\frac{6}{25}$256​.