Question 1050727: Having trouble with this homework!
Compute the probability of randomly drawing five cards from a deck and getting 3 Aces and 2 Kings. (Give your answer as a fraction.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you draw 5 cards from the deck.
the probability that you would get 3 aces and 2 kings would be calculated as follows:
the number of ways you can get 3 aces out of 4 aces is c(4,3) = 4.
the number of ways you can get 2 kings out of 4 kings is c(4,2) = 6.
the number of ways you can get 2 kings and 3 aces is 4 * 6 = 24.
the number of ways you can get 5 cards out of a deck of 52 cards is c(52,5) = 2598960.
the probability of getting 3 aces and 2 kings when you draw 5 cards from the deck is 24 / 2598960 = 9.234463016 * 10^-6.
there is another way to analyze it.
the answer should be the same.
the probability of getting 3 aces and 2 kings would be:
4/52 * 3/51 * 2/50 * 4/49 * 3/48 * 5! / (3! * 2!).
use your calculator to get 9.234463016 * 10^-6.
the answers are the same.
the two methods are good.
c(n,x) is equal to n! / (x! * (n-x)!).
this is the combination formula for the number of ways of getting x things out of n where order is not important.
if you want to get 3 aces out of 4 possible, the formula becomes c(4,3) = 4! / (3! * 1!) = (4*3*2*1) / (3*2*1*1) = 4.
if you want to get 2 kings out of 4 possible, the formula becomes c(4,2) = 4! / (2! * 2!) = (4*3*2*1) / (2*1*2*1) = 6.
the number of possible combinations of 3 aces and 2 kings becomes 4 * 6 = 24.
the number of ways you can get 5 cards, regardless of what they are, out of 52 cards is c(52,5) = 52! / (5! * 47!) = (52*51*50*49*48*47!) / (5*4*3*2*1*47!)/
the 47! in the numerator and denominator cancel out and you are left with (52*51*50*49*48) / (5*4*3*2*1) which is equal to 2598960.
24/2598960 = 9.234463016 * 10^-6 as we calculated before.
the second method finds the probability of of getting them in a specific order and then multiplies by the number of ways you can arrange them.
(4/52 * 3/51 * 2/50) is the probability of getting the 3 aces.
(4/49 * 4/48) is the probability of getting the 2 kings once you've drawn the 3 aces.
(4/52 * 3/51 * 2/50 * 4/49 * 4/48) is the probability of getting the 3 aces and the 4 kings in that specific order (first the 3 aces and then the 2 kings).
these 5 cards can be arranged in 5! ways.
however, 3 of them are the same and 2 of them are the same, therefore, the number of ways they can be arranged becomes 5! / (3! * 2!).
you are not distinguishing between the aces or the kings in anyway, such as suit, therefore they are assumed to be the same cards.
an ace is an ace regardless if it's a spade or a heart.
both ways are valid and useful to know.
i usually try to solve these both ways as a check against each method.
if the results are different, i try to find out why.
i hope you found this helpful.
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