AndreiGMAT wrote:Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

(A) 11/8, (B) 7/8, (C) 9/64, (D) 5/64, (E) 3/64

We are first given the individual probabilities that Xavier, Yvonne, and Zelda WILL solve the problem. We list these below:

P(Xavier will solve) = 1/4

P(Yvonne will solve) = 1/2

P(Zelda will solve) = 5/8

However, we see the question asks for the probability that Xavier and Yvonne, but not Zelda, will solve the problem.

Thus we must determine the probability that Zelda WILL NOT solve the problem. "Solving" and "not solving" are complementary events. When two events are complementary, knowing the probability that one event will occur allows us to calculate the probability that its complement will occur. That is, P(A) + P(Not A) = 1. In the case of Zelda, the probability that she WILL NOT solve the problem is the complement of the probability that she WILL solve the problem.

P(Zelda will solve) + P(Zelda will not solve) = 1

5/8 + P(Zelda will not solve) = 1

P(Zelda will not solve) = 1 - 5/8 = 3/8

Now we can determine the probability that Xavier and Yvonne, but not Zelda, will solve the problem. Since we need to determine three probabilities that all must take place, we must multiply the probabilities together. Thus, we have:

P(Xavier will solve) x P(Yvonne will solve) x P(Zelda will not solve)

1/4 x 1/2 x 3/8

1/8 x 3/8 = 3/64

Answer:

E