Problem Statement
Two cards are drawn simultaneously without replacement from a well-shuffled pack of 52 cards. Find the probability distribution and the mean of the number of aces.
Verified Solution & Marking Scheme
Define random variable X
Total cards $= 52$, Ace cards $= 4$, Non-ace cards $= 48$.
Let $X$ denote the number of aces drawn in 2 cards. Then $X$ can take values $0, 1, 2$.
Calculate probabilities for each value of X
$P(X = 0) = \frac{\binom{4}{0} \binom{48}{2}}{\binom{52}{2}} = \frac{1 \times \frac{48 \times 47}{2}}{\frac{52 \times 51}{2}} = \frac{48 \times 47}{52 \times 51} = \frac{12 \times 47}{13 \times 51} = \frac{4 \times 47}{13 \times 17} = \frac{188}{221}$
$P(X = 1) = \frac{\binom{4}{1} \binom{48}{1}}{\binom{52}{2}} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221}$
$P(X = 2) = \frac{\binom{4}{2} \binom{48}{0}}{\binom{52}{2}} = \frac{6 \times 1}{1326} = \frac{1}{221}$
Check sum: $\frac{188 + 32 + 1}{221} = \frac{221}{221} = 1$.
Compute the Mean (Expectation) E(X)
$\text{Mean } E(X) = \sum x_i P(X = x_i) = 0 \left(\frac{188}{221}\right) + 1 \left(\frac{32}{221}\right) + 2 \left(\frac{1}{221}\right) = \frac{32 + 2}{221} = \frac{34}{221} = \frac{2}{13}$