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
My flawed solution
Probability of successful outcomes
X-yes, Y-no, Z-no: 1/4*1/2*3/8=3/64
X-no, Y-yes, Z-no: 3/4*1/2*3/8=3/64
X-yes, Y-yes, Z-no: 1/4*1/2*3/8=3/64
Add them: 3/64+3/64+3/64=9/64
I used this method to solve problems with coin tosses and it worked. Not this time. Where is my mistake?
Probability - What's Wrong?
This topic has expert replies
- AndreiGMAT
- Junior | Next Rank: 30 Posts
- Posts: 18
- Joined: Tue Apr 05, 2016 3:31 am
- Followed by:1 members
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Only the case in blue -- yes for X, yes for Y, no for Z -- satisfies the constraint that Xavier AND Yvonne, but NOT Zelda, will solve the problem.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
My flawed solution
Probability of successful outcomes
X-yes, Y-no, Z-no: 1/4*1/2*3/8=3/64
X-no, Y-yes, Z-no: 3/4*1/2*3/8=3/64
X-yes, Y-yes, Z-no: 1/4*1/2*3/8=3/64
Add them: 3/64+3/64+3/64=9/64
I used this method to solve problems with coin tosses and it worked. Not this time. Where is my mistake?
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- AndreiGMAT
- Junior | Next Rank: 30 Posts
- Posts: 18
- Joined: Tue Apr 05, 2016 3:31 am
- Followed by:1 members
GMATGuruNY
You are so to the point!
The questions says: Xavier AND Yvonne, but not Zelda, not Xavier OR Yvonne.
You are so to the point!
The questions says: Xavier AND Yvonne, but not Zelda, not Xavier OR Yvonne.
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2, 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
First, 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
Related Resources
The following videos cover the concepts/strategies that are useful for answering this question:
- The Complement: https://www.gmatprepnow.com/module/gmat ... /video/744
- Probability of Event A AND Event B: https://www.gmatprepnow.com/module/gmat ... /video/750
- Rewriting Questions: https://www.gmatprepnow.com/module/gmat ... /video/754
- General Probability Strategies: https://www.gmatprepnow.com/module/gmat ... /video/757
Cheers,
Brent
GMAT/MBA Expert
- [email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi AndreiGMAT,
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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7245
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We are first given the individual probabilities that Xavier, Yvonne, and Zelda WILL solve the problem. We list these below: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
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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
- ygcrowanhand
- Senior | Next Rank: 100 Posts
- Posts: 30
- Joined: Mon Nov 24, 2014 1:47 pm
- Location: London
Here's my video solution to the problem:
https://youtu.be/Nwioau2P9hk
Additionally, there are many many more GMAT Probability resources available here:
https://privategmattutor.london/gmat-pr ... nd-videos/
Best,
Rowan
https://youtu.be/Nwioau2P9hk
Additionally, there are many many more GMAT Probability resources available here:
https://privategmattutor.london/gmat-pr ... nd-videos/
Best,
Rowan
Is Your GMAT Score Stuck in the 600s? This FREE 8-Video, 20-Page Guide Can Help.
https://privategmattutor.london/move-yo ... -the-700s/
PS have you seen the new GMAT Work and Rates guide? Comes with a free 8-video course.
https://yourgmatcoach.podia.com/courses ... s-problems
Learn more about Private GMAT Tutoring at: https://privategmattutor.london
https://privategmattutor.london/move-yo ... -the-700s/
PS have you seen the new GMAT Work and Rates guide? Comes with a free 8-video course.
https://yourgmatcoach.podia.com/courses ... s-problems
Learn more about Private GMAT Tutoring at: https://privategmattutor.london