Problem Statement
Two cards are drawn successively without replacement from a well-shuffled pack of $52$ cards. Find the probability distribution of the number of aces. Also, calculate the mean $E(X)$ of the distribution.
Verified Solution & Marking Scheme
Identify possible values of random variable X
Let $X$ denote the number of aces drawn. Since 2 cards are drawn, $X$ can take values $0, 1, 2$.
A standard pack has 4 aces and 48 non-aces.
Calculate probabilities for each value of X
$$P(X = 0) = \frac{48}{52} \times \frac{47}{51} = \frac{12}{13} \times \frac{47}{51} = \frac{564}{663} = \frac{188}{221}$$
$$P(X = 1) = \frac{4}{52} \times \frac{48}{51} + \frac{48}{52} \times \frac{4}{51} = 2 \left( \frac{1}{13} \times \frac{48}{51} \right) = \frac{96}{663} = \frac{32}{221}$$
$$P(X = 2) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$$
Check sum: $\frac{188 + 32 + 1}{221} = \frac{221}{221} = 1$.
Calculate Mean E(X) = ∑ x_i P(x_i)
$$E(X) = 0 \times \frac{188}{221} + 1 \times \frac{32}{221} + 2 \times \frac{1}{221} = \frac{0 + 32 + 2}{221} = \frac{34}{221} = \frac{2}{13}$$