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What is the probability that a randomly selected power hand contains examples 3 aces (event A) given that it contains 2aces (event B)​

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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%.

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