Bunuel wrote:

There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one blue marble and two red marbles from the bowl after three successive marbles are withdrawn from the bowl?

A. 2/81

B. 3/28

C. 2/27

D. 1/28

E. 1/84

We are given there are 3 blue marbles, 3 red marbles, and 3 yellow marbles in a bowl. We must determine the probability of selecting one blue and two red marbles in 3 attempts.

On the first draw, since there are 3 red marbles and 9 total marbles, there is a 3/9 chance that a red marble will be selected. Next, since there are 2 red marbles and 8 total marbles left, there is a 2/8 chance a red marble will be selected on the second draw. Finally, when selecting the blue marble, since there are 3 blue marbles and 7 marbles left, there is a 3/7 chance a blue marble will be selected for the final marble. However, there are 3 different ways to select the 2 red and 1 blue marbles:

R - R - B:

R - B - R:

B - R - R:

Note that each of these 3 ways has the same probability of occurring, even though the individual probabilities appear in a different order. Thus, the total probability is:

3 x (3/9 x 2/8 x 3/7) = 3 x (1/3 x 1/4 x 3/7) = 3/28

Alternate Solution:

There are 9C3 = (9 x 8 x 7)/(3 x 2 x 1) = 84 ways to choose 3 marbles from a total of 9 marbles.

There are 3C1 = 3 ways to choose a blue marble and 3C2 = 3 ways to choose a red marble. Thus, there are 3 x 3 = 9 ways to make a selection that involves two red and one blue marble.

Thus, the probability that the selection consists of two red marbles and one blue marble is 9/84 = 3/28.

Answer: B

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Jeffery Miller

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