Correct Option: B Let $E$ be the event that the hand contains at least two cards of the same rank. Let $F$ be the event that the hand contains exactly one Ace. We need to find $P(E|F) = \frac{P(E \cap F)}{P(F)}$. This can be calculated as $\frac{|E \cap F|}{|F|}$. **Step 1: Calculate the number of hands with exactly one Ace ($|F|$).** To form a 5-card hand with exactly one Ace: * Choose 1 Ace from the 4 Aces: $\binom{4}{1}$ ways. * Choose the remaining 4 cards from the 48 non-Ace cards: $\binom{48}{4}$ ways. $|F| = \binom{4}{1} \times \binom{48}{4} = 4 \times \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 4 \times 194580 = 778320$. **Step 2: Calculate the number of hands with exactly one Ace AND at least two cards of the same rank ($|E \cap F|$).** It's easier to calculate the complement: the number of hands with exactly one Ace AND all 5 cards have distinct ranks. Let this complement event be $E^c$. So, $|E \cap F| = |F| - |E^c \cap F|$. Now, let's find $|E^c \cap F|$ (exactly one Ace AND all 5 cards have distinct ranks): * Choose 1 Ace from 4: $\binom{4}{1}$ ways. (This fixes one card and its rank, which is Ace). * We need 4 more cards, none of which are Aces, and all of different ranks from each other AND different from the Ace's rank. There are 12 other ranks available (King, Queen, Jack, ..., 2). * Choose 4 distinct ranks from these 12 available ranks: $\binom{12}{4}$ ways. * For each of these 4 chosen ranks, select 1 card. Since the cards must have distinct ranks, and we are not picking an Ace, each of these 4 ranks has 4 possible suits to choose from. So, $4^4$ ways to choose the suits for these 4 cards. $|E^c \cap F| = \binom{4}{1} \times \binom{12}{4} \times 4^4$ $\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$ $4^4 = 256$ $|E^c \cap F| = 4 \times 495 \times 256 = 1980 \times 256 = 506880$. Now, $|E \cap F| = |F| - |E^c \cap F| = 778320 - 506880 = 271440$. **Step 3: Calculate the conditional probability $P(E|F)$.** $P(E|F) = \frac{|E \cap F|}{|F|} = \frac{271440}{778320}$ To simplify the fraction: $\frac{271440}{778320} = \frac{27144}{77832}$ Divide by 8: $\frac{3393}{9729}$ Divide by 3: $\frac{1131}{3243}$ Divide by 3 again: $\frac{377}{1081}$ Recognize that $377 = 13 \times 29$ and $1081 = 29 \times 37$. So, $\frac{377}{1081} = \frac{13 \times 29}{37 \times 29} = \frac{13}{37}$. The probability is $\frac{13}{37}$.