Question 478984: Ok....You helped answer a question previously and it really helped. Do you think that you can help to explain what I am looking at here and help to apply a formula...I am drawing a blank.
If you are dealt 4 cards from a shuffled deck of 52 cards, find the probability of getting three aces and one king.
I serioulsy don't know where to start. Thank you for your time.
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If you are dealt 4 cards from a shuffled deck of 52 cards, find the probability of getting three aces and one king.
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# of ways to draw 3 aces: 4C3 = 4
# of ways to draw 1 king: 4C1 = 4
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# of ways to draw 3 aces and 1 king: 4*4 = 16
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# of ways to draw 4 cards randomly: 52C4
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P(3 aces and one king) = 16/270725
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Cheers,
Stan H.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you are dealt 4 cards from a deck of 52
you want to know the probability that you get 3 aces and 1 king.
there are 4 aces in the deck and there are 4 kings in the deck.
this is a correction from the last time i issued the answer. i said 1 king in the deck when i meant 4 kings in the deck. the rest of the answer was good and assumed the 4 kings in the deck.
the probability of getting 3 aces and 1 king, without replacement, would be:
4/52 * 3/51 * 2/50 * 4/49 = .00001448 * 4C1 = .000059101
using combination theory, the equation would be:
(4C3 * 4C1) / (52C4) = (4*4)/270725 = 16/270725 = .000059101
both methods get you the same answer.
the difference from your other problems is the number of ways the probability can be done.
the probability method is:
4/52 * 3/51 * 2/50 * 4/49.
you are talking about 3 aces and 1 king.
this could be symbolized as aaak
you drew 4 cards out of the deck.
first one was ace, second one was ace, third one was ace, fourth one was king.
there are 3 other ways this same combination could have been drawn.
the 4 ways are:
aaak
aaka
akaa
kaaa
each one of these configurations has the same probability that is equal to 4/52 * 3/51 * 2/50 * 4/49.
to account for the fact that you could draw 3 aces and 1 king in 4 different ways, you have to multiply the probability by 4.
no replacement is assumed here so the 2 methods (probability methods and combination method) should be comparable as they are.
in the ticket problem, the number of ways was 1 because you were dealing with only tickets.
in the card problem, the number of ways was 1 because you were dealing with only clubs (same suit).
your set from the ticket problem was TT
your set from the club problem was CCC
either one of these can only happen in one way.
the problem with the ace and the king was AAAK which, as shown above, could be drawn in 4 ways.
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