Answer:
To find the conditional probability of event A given event B, we can use the formula:
P(A|B) = P(A ∩ B) / P(B)
where P(A ∩ B) is the probability that events A and B both occur, and P(B) is the probability of event B.
Assuming that a power hand consists of 5 cards, the probability that a randomly selected power hand contains exactly 2 aces (event B) can be calculated as follows:
P(B) = (4C2 * 48C3) / 52C5
= (6 * 17296) / 2598960
= 0.0404
where 4C2 is the number of ways to choose 2 aces from 4, 48C3 is the number of ways to choose 3 non-ace cards from the remaining 48 cards, and 52C5 is the total number of possible 5-card hands.
The probability that a randomly selected power hand contains exactly 3 aces (event A) and also contains 2 aces (event B) can be calculated as follows:
P(A ∩ B) = (4C3 * 48C2) / 52C5
= (4 * 1128) / 2598960
= 0.00173
where 4C3 is the number of ways to choose 3 aces from 4 (there is only one), and 48C2 is the number of ways to choose 2 non-ace cards from the remaining 48 cards.
Therefore, the conditional probability of event A given event B is:
P(A|B) = P(A ∩ B) / P(B)
= 0.00173 / 0.0404
= 0.0428
So the probability that a randomly selected power hand contains exactly 3 aces, given that it contains exactly 2 aces, is approximately 0.0428 or 4.28%.