Two cards are drawn **without replacement** from a standard $52$-card deck. What is the probability that **both** cards are aces?

First card ace: $P_{1} = \dfrac{4}{52} = \dfrac{1}{13}$. Given the first was an ace, only $3$ aces remain in $51$ cards: $P_{2 \mid 1} = \dfrac{3}{51} = \dfrac{1}{17}$. Combined: $P = \dfrac{1}{13} \cdot \dfrac{1}{17} = \dfrac{1}{221}$. - Trap A ($1/169 = (4/52)^2$) assumes replacement (independence). - Trap C is a wrong numerator. - Trap D divides wrong.