SOLUTION: An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles . If four marbles are drawn at random , what is the probability that two are blue and two are red ?
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Question 1120971: An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles . If four marbles are drawn at random , what is the probability that two are blue and two are red ? Answer by Edwin McCravy(20060) (Show Source):
The numerator of the desired probability is the number of ways
to choose two blues and two reds:
5 blue marbles choose 2 in 5C2 = (5∙4)/(2∙1) = 20/2 = 10 ways.
2 red marbles choose 2 in 2C2 = (2∙1)/(2∙1) = 2/2 = 1 way.
So the number of ways to choose 2 blues and 2 reds is:
(5C2)(2C2) = 10∙1 = 10 ways.
So the numerator of the desired probability is 10.
The denominator of the desired probability is the number of ways
to choose ANY 4 marbles:
4+5+2+3 = 14 marbles choose 4 = 14C4 = (14∙13∙12∙11)/(4∙3∙2∙1) = 24024/24 = 1001 ways.
So the denominator of the desired probability is 1001.
The desired probability is therefore 10/1001.
Edwin
