Problem Statement
Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards.
Verified Solution & Marking Scheme
Define Random Variable X and Probabilities
Let $X$ denote the number of red cards drawn. Possible values of $X$: $0, 1, 2$.
Total cards = 52 (26 red, 26 black).
$P(X = 0) = \frac{\binom{26}{0}\binom{26}{2}}{\binom{52}{2}} = \frac{325}{1326} = \frac{25}{102}$
$P(X = 1) = \frac{\binom{26}{1}\binom{26}{1}}{\binom{52}{2}} = \frac{26 \times 26}{1326} = \frac{676}{1326} = \frac{52}{102}$
$P(X = 2) = \frac{\binom{26}{2}\binom{26}{0}}{\binom{52}{2}} = \frac{325}{1326} = \frac{25}{102}$
Calculate Mean E(X)
$E(X) = \sum x_i P(x_i) = 0\left(\frac{25}{102}\right) + 1\left(\frac{52}{102}\right) + 2\left(\frac{25}{102}\right) = \frac{52 + 50}{102} = \frac{102}{102} = 1$
Calculate E(X²) and Variance
$E(X^2) = \sum x_i^2 P(x_i) = 0^2\left(\frac{25}{102}\right) + 1^2\left(\frac{52}{102}\right) + 2^2\left(\frac{25}{102}\right) = \frac{52 + 100}{102} = \frac{152}{102} = \frac{26}{17} \approx 1.490$
$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{26}{17} - 1^2 = \frac{9}{17} \approx 0.529$