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Probability that X and Y solve

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by LulaBrazilia » Mon Mar 17, 2014 8:30 am

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Answers

A

B

C

D

E

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Difficulty—

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
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by Patrick_GMATFix » Mon Mar 17, 2014 8:42 am
To solve we need to understand 2 things:
> Prob(success) + Prob(failure) = 1 for each individual
> Prob(independent events A and B both happen) = Prob(A) * Prob(B)

The answer is E. I go through the question in detail in the full solution below (taken from the GMATFix App).

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by [email protected] » Tue Mar 18, 2014 12:26 am
Hi LulaBrazilia,

In this type of probability question, we're asked for a specific outcome. To solve this problem, we'll have to deal with each piece individually, then multiply the outcomes together.

We're asked for 3 things:

Xavier solves the problem
Yvonne solves the problem
Zelda does NOT solve the problem.

Xavier's probability to solve = 1/4
Yvonne's probability to solve = 1/2
Zelda's probability to NOT solve = 1 - 5/8 = 3/8

The final answer is (1/4)(1/2)(3/8) = 3/64

Final Answer: E

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by Scott@TargetTestPrep » Tue Jul 28, 2015 1:59 pm
LulaBrazilia 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
Solution:

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

P(Xavier will solve) = ¼

P(Yvonne will solve) = ½

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 first 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 the 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 events that all must take place, we multiply their probabilities together. Thus, we have:

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

¼ x ½ x 3/8

1/8 x 3/8 = 3/64

The answer is E

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by Brent@GMATPrepNow » Tue Jul 28, 2015 2:11 pm
LulaBrazilia 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
P(Z solves the problem) = 1 - P(Z doesn't solve the problem)
So, 5/8 = 1 - P(Z doesn't solve the problem)
So, P(Z doesn't solve the problem) = 3/8

The question asks us to find P(Xavier and Yvonne solve problem, but Zelda does not solve problem)
So, we want: P(X solves problem AND Y solves problem AND Z does not solve)
= P(X solves problem) x P(Y solves problem) x P(Z does not solve)
= 1/4 x 1/2 x 3/8
= 3/64
= E

Cheers,
Brent
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by nikhilgmat31 » Tue Jul 28, 2015 11:57 pm
1/4 * 1/2 * (1-5/8)

1/8 * 3/8

=3/64
E
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