Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

I read the tutorial this forum offers, the NOT tool says:
p(not A) + p(A) = 1. I tried it but I could not come up with the answer. Thank you for helping.

The probability of Xavier solve the problem is 1/4
the probability of Yvonne solve the problem is 1/2
the probability of Zelda not solve the problem is 1-5/8, or 3/8
So the probability of Xavier and Yvonne but not Zelda solve the problem will be 1/4*1/2*3/8, or 3/64. Answer E

What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64

Is this right? I calculated it as:

Prob of X or Y= (1/4)+(1/2)-(1/4)(1/2)=(6/8)-(1/8)=5/8. Multiply by the probability that Zelda doesn't solve: (5/8)*(3/8)=15/64.

I think you just added 1, 2, and 3 incorrectly.

I dont think you can subtract (1/4) * (1/2). The assumption here is independent events. I may be wrong as I am having a horrible day today

dahcrap, dont beat up on yourself so much - you hav solved so many tough problems - i surely think u'll score in the 99th+ percentile.

anywya, my understanding is just the opposite
firstly, that you subtract Px + Py by Pxy bcoz they are independent (whereas, if they are mutually exclusive, there is nothing to subtract)
in other words, any two or all of them can solve the problem.

solution of asaf and briks seems correct to me (but of course, i am not an expert).

What is the probability of Xavier or Yvonne but not Zelda solving the problem?

1. X & Y but not Z = 1/4*1/2 *3/8 = 3/64

2. X and not Y and not Z = 1/4 * 1/2 * 3/8 = 3/64

3. not X and Y and not Z = 3/4*1/2*3/8 = 9/64

Adding 1, 2 and 3 = 13/64

Is this right? I calculated it as:

Prob of X or Y= (1/4)+(1/2)-(1/4)(1/2)=(6/8)-(1/8)=5/8. Multiply by the probability that Zelda doesn't solve: (5/8)*(3/8)=15/64.

I think you just added 1, 2, and 3 incorrectly.

I dont think you can subtract (1/4) * (1/2). The assumption here is independent events. I may be wrong as I am having a horrible day today

dahcrap, dont beat up on yourself so much - you hav solved so many tough problems - i surely think u'll score in the 99th+ percentile.

anywya, my understanding is just the opposite firstly, that you subtract Px + Py by Pxy bcoz they are independent (whereas, if they are mutually exclusive, there is nothing to subtract) in other words, any two or all of them can solve the problem.

solution of asaf and briks seems correct to me (but of course, i am not an expert).

on the other hand, in the case that they are mutually exclusive, the probability for X and Y are added, but P for Z is subtracted instead of multiplying by NOT Z.

for example, here is a question from gmatclub Challenge 7:
Probability Mike can win a championship is 1/4, Rob 1/3 and Ben 1/6. What is the Probability that Mike or Rob win but not Ben?

the answer to this from my notes: 1/4 + 1/3 - 1/6 = 5/12
my understanding is this is bcoz only one of them can win the championship, unlike the previous problem.

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?