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A poker hand, consisting of 9 cards, is dealt from a standard deck of 52 cards. Find the probability that the hand contains 9 hearts.

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Answer:

8.93 × 10^-6

Explanation:

The probability of getting a heart on the first draw is 13/52 since there are 13 hearts in a standard deck of 52 cards. After one heart has been drawn, there are 12 hearts left in the remaining 51 cards, so the probability of getting a heart on the second draw is 12/51. Similarly, the probability of getting a heart on the third draw is 11/50, and so on.

The probability of getting 9 hearts in a row is the product of the probabilities of getting a heart on each of the 9 draws:

(13/52) * (12/51) * (11/50) * (10/49) * (9/48) * (8/47) * (7/46) * (6/45) * (5/44)

Simplifying this expression, we get:

(13! / 4!)(40! / 31!)(31! / 22!)(22! / 13!)(13! / 4!)(8! / 1!)(7! / 1!)(6! / 1!)(5! / 1!) / 52!

where ! denotes factorial, the product of all positive integers up to and including that number. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120.

Canceling out the common terms, we get:

(13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5) / (52 * 51 * 50 * 49 * 48 * 47 * 46 * 45 * 44)

Simplifying further, we get:

1287 / 144074475

Therefore, the probability of getting a poker hand consisting of 9 hearts is approximately 0.00000893 or 8.93 × 10^-6.

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