Probability - What's Wrong?

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Probability - What's Wrong?

by AndreiGMAT » Wed Apr 27, 2016 10:42 pm

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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?

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by GMATGuruNY » Wed Apr 27, 2016 11:05 pm
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?
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.
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by AndreiGMAT » Thu Apr 28, 2016 12:32 am
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You are so to the point!
The questions says: Xavier AND Yvonne, but not Zelda, not Xavier OR Yvonne.

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by [email protected] » Thu Apr 28, 2016 6:17 am
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
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Thu Apr 28, 2016 9:20 am
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

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by [email protected] » Tue Jan 09, 2018 10:31 am
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

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Re: Probability - What's Wrong?

by ygcrowanhand » Tue May 24, 2022 6:11 am
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
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