SOLUTION: If five cards are drawn from a standard deck without replacement, what is the probability of drawing:
A full house (a three of a kind and a pair?)
I'm not sure if it shou
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-> SOLUTION: If five cards are drawn from a standard deck without replacement, what is the probability of drawing:
A full house (a three of a kind and a pair?)
I'm not sure if it shou
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Question 694933: If five cards are drawn from a standard deck without replacement, what is the probability of drawing:
A full house (a three of a kind and a pair?)
I'm not sure if it should be:
(52/52)*(3/51)*(2/50)*(48/49)*(3/48)*5C3*5C2
or just:
(52/52)*(3/51)*(2/50)*(48/49)*(3/48)*5C3
You can put this solution on YOUR website! If five cards are drawn from a standard deck without replacement, what is the probability of drawing:
A full house (a three of a kind and a pair?)
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Select a card: 13 ways
Select 3 of those cards: 4C3 = 4
Select a different card: 12 ways
Select 2 of those cards: 4C2 = 6
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Total # of ways to succeed: 13*4*12*6 = 2744
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Total # of 5 card sets: 52C5 = (52*51*50*49*48)/(1*2*3*4*5) = 2598960
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P(full house) = 2744/2598960 = 0.00106
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Your Questions:
5C3 means the number of groups of size 3 in a set of size 5
5C3 = (5*4*3)/(1*2*3) = 10
5C2 means the number of groups of size 2 in a set of size 5
5C2 = (5*4)/(1*2) = 10
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Why are the number's the same?
Each time you select 3 from 5 you are automatically selecting a group of 2.
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Cheers,
Stan H.
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