Direction: Study the given information carefully and answer the question that follow.
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.
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If three marbles are picked at random, what is the probability that two are blue and one is yellow ?
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- 3/91
- 1/5
- 18/455
- 7/15
Correct Option: C
Ways of selection of two blue marbles = 4C2
Ways of selection of one yellow marble = 3C1
Ways of selection of three marbles = 15C3
So, required probability = (4C2 x 3C1) / (15C3)
= [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}]
= [(4 x 3 x 2 x 1) / (2 x 1 x 2 x 1)] x [(3 x 2 x 1) / (1 x 2 x 1)] / [ (15 x 14 x 13 x 12!) / (3 x 2 x 12!)]
= (6 x 3) / (5 x 7 x 13)
= 18 / 455
