SOLUTION: an urn contains 4 red, 2 white, and 4 blue marbles. In how many ways can three marbles be selected so that at least one of them is blue?

Question 221752: an urn contains 4 red, 2 white, and 4 blue marbles. In how many ways can three marbles be selected so that at least one of them is blue?
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
The direct, straightforward way to solve this is to find the probability of exactly 1 blue marble, exactly 2 blue marbles and exactly 3 blue marbles and then add these probabilities together for your answer. This is not easy.

It is easier to realize that the probability of at least one blue marble is (1 - probability of zero blue marbles)! So all we need to figure out is the probability of zero blue marbles.

With the first selection the probability of not selecting a blue marble: 6/10 = 3/5.
With the second selection the probability of not selecting a blue marble (given that no blue marble has been selected earlier): 5/9.
With the third selection the probability of not selecting a blue marble (given that no blue marble has been selected earlier): 4/8 = 1/2.
The probability of all three:

So the probability of picking at least one blue = 1 - 1/6 = 5/6.
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