Two cards are drawn simultaneously from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards.

Mean = 1, Variance = $\frac{25}{51}$

The correct answer is Option (4) → Mean = 1, Variance = $\frac{25}{51}$

Let random variable X denote the number of red cards drawn in a draw of 2 cards from a pack of 52 cards, then X can take values 0, 1, 2.

$\text{P(X = 0) = P(2 black cards)} = \frac{{^{26}C}_2}{{^{52}C}_2}=\frac{26×25}{1×2}+\frac{1×2}{52×51}=\frac{25}{102}$,

$\text{P(X = 1) = P(1 red card and 1 black card)} =\frac{{^{26}C}_1×{^{26}C}_1}{{^{52}C}_2}=\frac{26×26×1×2}{52×51}=\frac{52}{102}$,

$\text{P(X = 2) P(2 red cards)} =\frac{{^{26}C}_2}{{^{52}C}_2}=\frac{26×25}{1×2}+\frac{1×2}{52×51}=\frac{25}{102}$

We construct the table as under:

∴ Mean = $Σp_ix_i = 1$.

Variance = $Σp_i{x_i}^2  - μ^2 =\frac{152}{102}-(1)^2=\frac{50}{102}=\frac{25}{51}$