A box contains 1 blue ball, 1 green ball, 1 yellow ball

Gmat_mission wrote:A box contains 1 blue ball, 1 green ball, 1 yellow ball, and 2 red balls. Three balls are randomly selected (one after the other) without replacement. What is the probability that the 2nd ball is NOT red and the 3rd ball is yellow?

A) 1/30
B) 1/20
C) 1/10
D) 3/20
E) 1/5

Since there is only one yellow ball and the selection is without replacement, neither the first nor the second ball can be yellow if the third one must be yellow. So the first ball can be the blue ball, the green ball, or one of the two red balls. Let’s say the first ball is blue or green; then we have:

P(1st ball is blue or green) x P(2nd ball is neither yellow nor red) x P(3rd ball is yellow)

2/5 x 1/4 x 1/3

2/60

Now, let’s say the first ball is red; then we have:

P(1st ball is red) x P(2nd ball is not yellow or red) x P(3rd ball is yellow)

2/5 x 2/4 x 1/3

4/60

Thus the overall probability is 2/60 + 4/60 = 6/60 = 1/10.

Alternate Solution:

Let’s interpret the selections as orderings of the letters B, G, Y and R. Without any restrictions, 3 of these 4 letters can be ordered in 5P3 = 5!/(5-3)! = 5 x 4 x 3 = 60 ways.

Let’s list all the orderings where the second letter is not R and the third letter is Y. We have two cases of RGY (one for each red ball), two cases of RBY, one case of BGY and one case of GBY. In total, there are 6 cases where the second letter is not red and the third letter is yellow.

Thus, the probability of a selection with the desired properties is 6/60 = 1/10.

Answer: C