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02 Jan 2021, 10:54
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (02:45) correct
53%
(03:16)
wrong
based on 150
sessions
History
Date
Time
Result
Not Attempted Yet
Two canoe riders must be selected from each of two groups of campers. 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 two men and two 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
Shadyshades
Joined: 02 Nov 2020
Last visit: 21 Apr 2026
Posts: 113
Given Kudos: 1,150
Location: India
Schools: Yale '20 (D)
GMAT 1: 220 Q2 V2
02 Jan 2021, 12:37
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Conditions-
I have to take two members from each group.
AND
I have to take 2 Men and 2 Women in total.
Group A has 3 Men and 1 Woman
Group B has 1 Man and 2 Women
Case 1 - I can choose 2 Men from Group A and 2 Women from Group B
so the number of combinations are 3C2*2C2 = 3*1= 3
Case 2 - I can choose 1 Man 1 Woman from Group A and 1 Man 1 Woman from Group B
so the number of combinations are 3C1*1C1*1C1*2C1 = 3*1*1*2= 6
Total number of combinations are 4C2*3C2= 6*3= 18
(Case 1 + Case 2) / Total combinations
=(6+3)/18
=9/18
=1/2
Option E
I have to take two members from each group.
AND
I have to take 2 Men and 2 Women in total.
Group A has 3 Men and 1 Woman
Group B has 1 Man and 2 Women
Case 1 - I can choose 2 Men from Group A and 2 Women from Group B
so the number of combinations are 3C2*2C2 = 3*1= 3
Case 2 - I can choose 1 Man 1 Woman from Group A and 1 Man 1 Woman from Group B
so the number of combinations are 3C1*1C1*1C1*2C1 = 3*1*1*2= 6
Total number of combinations are 4C2*3C2= 6*3= 18
(Case 1 + Case 2) / Total combinations
=(6+3)/18
=9/18
=1/2
Option E
13 Feb 2021, 23:25
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Bunuel
The crux of the question is that two riders must be selected from each of the two groups. Had this not been the case, the four riders can be selected as usual in \(7_{C_4}\) ways.
Required selection can be done in ways:
1. For 1st group(3M,1W) - \(4_{C_2}\)
2. For 2nd group(2W,1M) - \(3_{C_2}\)
Hence, \(\frac{4*3}{2}*\frac{3*2}{2} = 18\)
Now, 2M and 2W can be selected in ways:
A) 2M from 1st group and 2W from 2nd group - \(3_{C_2}*2_{C_2} = 3\)
B) 1M and 1W from each group - \(3_{C_1}*1_{C_1}*2_{C_1}*1_{C_1} = 6\)
P(2M,2W) = \(\frac{3+6}{18} = \frac{1}{2}\)
Answer E.
