(N/A) Let $X$ be the random variable representing the number of aces obtained in two draws with replacement. The possible values for $X$ are $0, 1, 2$.
The probability of drawing an ace in a single draw is $p = \frac{4}{52} = \frac{1}{13}$.
The probability of not drawing an ace is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since the draws are independent (with replacement),we use the binomial distribution $P(X=k) = \binom{n}{k} p^k q^{n-k}$ where $n=2$:
$P(X=0) = \binom{2}{0} (\frac{1}{13})^0 (\frac{12}{13})^2 = 1 \times 1 \times \frac{144}{169} = \frac{144}{169}$
$P(X=1) = \binom{2}{1} (\frac{1}{13})^1 (\frac{12}{13})^1 = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169}$
$P(X=2) = \binom{2}{2} (\frac{1}{13})^2 (\frac{12}{13})^0 = 1 \times \frac{1}{169} \times 1 = \frac{1}{169}$
The probability distribution is:
| $X$ | $P(X)$ |
| $0$ | $\frac{144}{169}$ |
| $1$ | $\frac{24}{169}$ |
| $2$ | $\frac{1}{169}$ |