Explanation

(c)
Let $X$ denote the number of aces.
Probability of selecting aces,
$P = \frac{4}{52} = \frac{1}{13}$
Probability of not selecting aces,
$q = 1 - \frac{1}{13} = \frac{12}{13}$
$P(X = 1) = 2 \times \left( \frac{1}{13} \right) \times \left( \frac{12}{13} \right) = \frac{24}{169}$
$P(X = 2) = 2\left( \frac{1}{13} \right)^{2}.\left( \frac{12}{13} \right)^{0} = \frac{2}{169}$
Mean $= \Sigma P_{1}X_{i} = \frac{24}{169} + \frac{2}{169} = \frac{2}{13}$