Probability OG 13 Q 215

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rchetan_18: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sat Dec 01, 2012 6:47 am

by rchetan_18 » Sat Feb 02, 2013 6:10 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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand
Here's another approach -
Zelda's failure = 3/8 = probability the question is solved by Xavier or Yvonne - IS THIS CORRECT?
so using the probability OR formula
3/8 = 1/4 + 1/2 - P(Xavier and Yvonne)
P (X&Y) = 6/8 - 3/8 = 3/8
What am i doing wrong?

Thanks in advance for your help

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Source: Beat The GMAT — Problem Solving | Sat Feb 02, 2013 6:10 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Feb 02, 2013 6:30 am

rchetan_18 wrote: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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand
Here's another approach -
Zelda's failure = probability the question is solved by Xavier or Yvonne - IS THIS CORRECT?

Thanks in advance for your help

The portion in red is incorrect.
Xavier, Yvonne and Zelda each try INDEPENDENTLY to solve the problem.
The probability that Zelda doesn't solve the problem does NOT depend on whether Xavier or Yvonne solves the problem.
It is entirely possible that ALL 3 solve the problem.
Thus, it is not necessarily true that P(Z doesn't solve the problem) = P(X or Y solves the problem).

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Feb 02, 2013 7:19 am

rchetan_18 wrote: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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand

I thought I'd elaborate on your solution, so that others understand what you did here.

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Fazze: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sat Dec 22, 2012 8:46 am

by Fazze » Mon Feb 04, 2013 3:09 am

this seemed easy... can any one tell me which level is this ... below 600 ? :/

finished it in 49 seconds... all the probability is given...

it's just P {X & Y & (not Z)}
= 1/4 * 1/2 * (1-5/8)
=1/4 * 1/2 * 3/8

3/64

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shaan17: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Thu Aug 02, 2012 1:04 am

by shaan17 » Mon Apr 22, 2013 1:12 am

Hi,

When I solved this problem I multiplied the result with 3 to consider the 3 different sequences in which this task can be done. But I now realize that the different orders are not required. How is this problem different from other similar problems which require us to do so ? Is it correct to say that for ALL independent events we do not consider the order ?

Consider the example where we have 3 Blue balls and 2 Green balls. If I have to take 2 Blue and 1 Green balls at random then the probability is 3/5 x 2/4 + 2/5 x 3/4.
Is it because this scenario has dependent events ?

Thanks
Karthik

GMATGuruNY wrote:
rchetan_18 wrote: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

The solution is 1/4 * 1/2 * 3/8 (Zelda's failure) which I understand
Here's another approach -
Zelda's failure = probability the question is solved by Xavier or Yvonne - IS THIS CORRECT?

Thanks in advance for your help
The portion in red is incorrect.
Xavier, Yvonne and Zelda each try INDEPENDENTLY to solve the problem.
The probability that Zelda doesn't solve the problem does NOT depend on whether Xavier or Yvonne solves the problem.
It is entirely possible that ALL 3 solve the problem.
Thus, it is not necessarily true that P(Z doesn't solve the problem) = P(X or Y solves the problem).

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