BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 69
Joined: Wed Aug 08, 2012 9:00 am
Followed by:1 members

Probability question

by sanaa.rizwan » Mon Apr 08, 2013 4:41 pm
NP: OG PS 215

X, Y & Z each try to independently solve a problem, if their individual probabilities for success are 1/4, 1/2 and 5/8 respectively, what is the probability that X & Y but not will solve the problem

A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 511
Joined: Wed Aug 11, 2010 9:47 am
Location: Delhi, India
Thanked: 344 times
Followed by:86 members

by Anju@Gurome » Mon Apr 08, 2013 6:55 pm
sanaa.rizwan wrote:X, Y & Z each try to independently solve a problem, if their individual probabilities for success are 1/4, 1/2 and 5/8 respectively, what is the probability that X & Y but not Z will solve the problem
I think the part in blue was missing from your post.

Required probability = X and Y solves the problem but Z doesn't
= (Probability that X solves the problem)*(Probability that Y solves the problem)*(Probability that Z doesn't solve the problem)
= (1/4)*(1/2)*(1 - 5/8)
= (1/4)*(1/2)*(3/8)
= 3/64

The correct answer is E.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Top
Post new topic Post Reply
• Page 1 of 1