Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 1807:30 AM PDT
-08:30 AM PDT
Join live on GMAT Club's YouTube channel as we hear his success story after getting Rejected by ISB.
May 1601:30 PM IST
-11:00 PM IST
With brand new features like:AI-driven Planner tool, 850+ data Insights practice questions and GMAT Focus Edition Adaptive mock tests with ESR+ analysis and personal mentor support, our course is the most comprehensive course for GMAT Focus Edition.
May 1805:30 AM PDT
-07:30 AM PDT
Unlock GMAT success! Join our webinar for exclusive insights on scoring V85+ in the Verbal section. Proven strategies, expert tips, and essential focus to elevate your GMAT game. Register now for the ultimate Verbal mastery!- May 19
07:30 AM PDT
-08:30 AM PDT
Solve 30 GMAT Focus practice problems covering Problem Solving, Data Sufficiency, Data Insights, and Critical Reasoning. Take this GMAT practice quiz live with peers, analyze your GMAT study progress, and see where you stand in the GMAT student pool. - May 19
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. - May 20
12:00 PM PDT
-01:00 PM PDT
Get Ready for a Truly Personalized Testing Experience. - May 21
07:30 AM PDT
-08:30 AM PDT
Are you curious about the dynamic field of Product Management and how it can shape your career trajectory? Join Carnegie Mellon for an insightful conversation about the MS in Product Management (MSPM) ...
Two representatives must be selected from each of two groups of studen
[#permalink]
07 Dec 2010, 06:42
07 Dec 2010, 06:42
3
Bookmarks
Show timer
00:00
A
B
C
D
E
Difficulty:
75%
(hard)
Question Stats:
47% (02:14) correct
53%
(02:45)
wrong
based on 63
sessions
Two representatives must be selected from each of two groups of students. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that 2 men and 2 women will be selected ?
A. 1/6
B. 1/4
C. 2/7
D. 1/3
E. 1/2
A. 1/6
B. 1/4
C. 2/7
D. 1/3
E. 1/2
Re: Two representatives must be selected from each of two groups of studen
[#permalink]
07 Dec 2010, 07:09
07 Dec 2010, 07:09
Expert Reply
tomchris wrote:
two representatives must be selected from each of two groups of students . one group consists of three men and one woman ,and the other group consists of two women and one man . what is the probability that 2 men amd 2 women will be selected ?
a) 1/6
b) 1/4
c) 2/7
d) 1/3
e) 1/2
a) 1/6
b) 1/4
c) 2/7
d) 1/3
e) 1/2
Group #1: {m1, m2, m3, w1};
Group #2: {m4, w2, w3}.
There are following cases possible to select 2m and 2w:
Both men from group #1 and both women from group #2: \(C^2_3*C^2_2=3\);
One man and one women from each group: \((C^1_3*C^1_1)*(C^1_1*C^1_2)=6\);
Total ways to select 2 persons from group #1 and group #2 is \(C^2_4*C^2_3=18\);
\(P=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}=\frac{3+6}{18}=\frac{1}{2}\).
Answer: E.
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 5983
Given Kudos: 124
Location: India
GMAT: QUANT+DI EXPERT
GMAT 1: 750 Q51 V41
WE:Education (Education)
Re: Two representatives must be selected from each of two groups of studen
[#permalink]
05 Feb 2016, 11:28
05 Feb 2016, 11:28
Expert Reply
tomchris wrote:
Two representatives must be selected from each of two groups of students. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that 2 men and 2 women will be selected ?
A. 1/6
B. 1/4
C. 2/7
D. 1/3
E. 1/2
A. 1/6
B. 1/4
C. 2/7
D. 1/3
E. 1/2
Group 1: 3 Men and 1 Woman
Group 2: 1 Men and 2 Woman
Need: 2 Men and 2 women
Case 1: 2 Men from Group 1 and 2 women from group 2 - 3C2*2C2 = 3 ways
Case 2: 1 Men and 1 Woman from Group 1 and 1 Man and 1 women from group 2 - 3*1*1*2 = 6 ways
Total Favorable cases = 3+6 = 9
Total Possible ways of selecting students 2 from each group = 4C2*3C2 = 6*3 = 18
Probability = 9/18 = 1/2
Answer : Option E
