An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. — Answer & Discussion

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.
If three marbles are picked at random, what is the probability that two are blue and one is yellow?

A3/91
B1/5
C18/455
D7/15

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Discussion 2

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CCommunity09 Apr 2019

Solution
Number of ways of selecting 2 blue urns and 1 yellow urn = $^{4}C_2* ^{3}C_1$ = 18
Probability of selecting 2 blue urns and 1 yellow urn = $\frac{18}{^{15}C_3}$ = $\frac{18}{455}$

CCommunity02 May 2019

Solution
Number of ways of selecting 2 blue urns and 1 yellow urn = $^{4}C_2* ^{3}C_1$ = 18
Probability of selecting 2 blue urns and 1 yellow urn = $\frac{18}{^{15}C_3}$ = $\frac{18}{455}$