Suppose two cards are drawn at random from a deck of cards, Let \( X \) be the number of aces obtained. Then the value of \( \mathrm{E}(\mathrm{X}) \) is
Let \(X = X_1 + X_2\), where \(X_i = 1\) if the \(i\)-th card drawn is an ace.
Each card is equally likely to be any card, so \(E(X_i) = \frac{4}{52} = \frac{1}{13}\).
By linearity of expectation, \(E(X) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}\).
(Check with the distribution: \(P(X=1) = \frac{4 \cdot 48}{\binom{52}{2}} = \frac{192}{1326}\), \(P(X=2) = \frac{6}{1326}\). Then \(E(X) = \frac{192 + 12}{1326} = \frac{204}{1326} = \frac{2}{13}\).)
Answer: (D) 2/13
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