Question 1119661: A poker hand consists of five cards randomly dealt from a standard deck of 52 cards. The order of the cards does not matter. Determine the following probabilities for a 5-card poker hand. Write your answers in percent form, rounded to 4 decimal places.
a) Determine the probability that exactly 3 of these cards are Aces.
b) Determine the probability of selecting exactly 2 Aces and exactly 2 Kings
c) Determine the probability of selecting exactly 1 Jack.
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
These are all straightforward probability questions that can be answered using the "n choose r" concept and basic rules of probability.
The number of possible hands is "52 choose 5":
C(52,5) = 2598960.
(a) To get exactly 3 aces, you need to choose 3 of the 4 aces and 2 of the other 48 cards. The number of ways to do that is
C(4,3)*C(48,2) = 6768
The probability of getting exactly 3 aces is then 6768/2598960 = .0026
(b) To get exactly 2 aces and 2 kings, you need to choose 2 of the 4 aces, 2 of the 4 kings, and 1 of the other 44 cards. The number of ways to do that is
C(4,2)*C(4,2)*C(44,1) = 1584
The probability is then 1584/2598960 = .00006 (.0001, to 4 decimal places)
(c) To get exactly 1 jack, you need to choose 1 of the 4 jacks and 4 of the other 48 cards. The number of ways to do that is
C(4,1)*C(48,4) = 778320
The probability is then 778320/2598960 = .2995
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Greenestamps has the right idea, but made a few typos.
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Part (a)
C(4,3)*C(48,2) = 4512 instead of 6768
The answer should be 4512/(2,598,960) = 0.0017
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Part (b)
The answer is 1584/(2,598,960) = 0.0006 (note there are three zeros between the decimal point and the 6; not four zeros)
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Part (c)
The answer he wrote for this part is correct.
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