both Peter and Paul win

This topic has expert replies
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

both Peter and Paul win

by sanju09 » Fri Jul 04, 2014 10:07 pm
The probability that Peter would win a grill show is twice of the probability that Paul would win the same grill show. If the probability that both Peter and Paul win the grill show is 9/32, what is the probability that only one of the two would win the grill show?
A. ¾
B. 9/16
C. 3/8
D. ¼
E. 5/32


Made Up!
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Source: — Problem Solving |

User avatar
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

by GMATGuruNY » Sat Jul 05, 2014 3:34 am
sanju09 wrote:The probability that Peter would win a grill show is twice of the probability that Paul would win the same grill show. If the probability that both Peter and Paul win the grill show is 9/32, what is the probability that only one of the two would win the grill show?
A. ¾
B. 9/16
C. 3/8
D. ¼
E. 5/32
Let P(Paul wins) = x.
Since Peter's probability is twice Paul's, P(Peter wins) = 2x.
Since the probability that both win is 9/32, we get:
x * 2x = 9/32.
2x² = 9/32
x² = 9/64
x = 3/8.

Thus:
P(Paul wins) = x = 3/8, implying that P(Paul loses) = 5/8.
P(Peter wins) = 2x = 2 * (3/8) = 3/4, implying that P(Peter loses) = 1/4.

Case 1: P(Paul wins but Peter doesn't) = 3/8 * 1/4 = 3/32.
Case 2: P(Peter wins but Paul doesn't) = 3/4 * 5/8 = 15/32.
Since exactly one person wins in either Case 1 or Case 2, we ADD the fractions:
3/32 + 15/32 = 18/32 = 9/16.

The correct answer is B.
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

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Sun Jul 06, 2014 6:21 pm
You might want to stipulate that the two men's chances of winning are independent, as the GMAC has been dabbling with dependent probability of late, at least going by the Question Pack they just released. (Probability in general seems to be a bigger focus of the test, though the new OG will tell us more.)

Anyway, I thought of this problem as:

(Prob ONLY One Wins) = 1 - (Both Win) - (Neither Wins)
= 1 - (x*2x) - (1-x)(1-2x)
= 3x - 4x²
= x * (3 - 4x)

We're given that 2x² = 9/32, or x = 3/8, so we have

(3/8) * (3 - 3/2), or (3/8) * (3/2), or 9/16.

User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Mon Jul 14, 2014 11:27 pm
Great replies!

If P(Paul wins) = x then P(Peter wins) = 2x; such that (x)(2x) = 9/32 or x = 3/8 (probability that Paul loses is 5/8) and 2x = ¾ (probability that Peter loses is ¼).

Now, the probability that only one of the two would win the grill show means

Peter wins and Paul loses OR Paul wins and Peter loses

= (¾) (5/8) + (3/8)( ¼)

= (15/32) + (3/32)

[spoiler]= 18/32

= 9/16.

Chose (B).
[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com