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24 Apr 2021, 06:42
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (01:52) correct
30%
(02:12)
wrong
based on 162
sessions
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Not Attempted Yet
There are two sets of dancers and a single pair must be randomly selected from each set. The first set consists of three men and one woman, and the second set consists of two women and one man. What is the probability that two men will be selected from the first set and two women will be selected from the second set?
(A) 1/8
(B) 1/6
(C) 2/7
(D) 1/3
(E) 1/2
(A) 1/8
(B) 1/6
(C) 2/7
(D) 1/3
(E) 1/2
Updated on: 22 Feb 2025, 04:19
Originally posted by GMATinsight on 24 Apr 2021, 08:12.
Last edited by Bunuel on 22 Feb 2025, 04:19, edited 2 times in total.
Last edited by Bunuel on 22 Feb 2025, 04:19, edited 2 times in total.
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Bunuel
The first set consists of three men and one woman, i.e. Total members = 3+1= 4
Total ways of choosing 2 members out of 4 = \(^4C_2 = 6\)
Ways of selecting 2 men from first set = \(^3C_2 = 3\)
Probability of selecting 2 men from first set = \(\frac{3}{6}\)
the second set consists of two women and one man, i.e. Total members = 2+1= 3
Total ways of choosing 2 members out of 3 = \(^3C_2 = 3\)
Ways of selecting 2 Women from Second set = \(^2C_2 = 1\)
Probability of selecting 2 women from first set = \(\frac{1}{3}\)
Required Probability = \(\frac{3}{6}\)*\(\frac{1}{3}\) = \(\frac{1}{6}\)
Answer: Option B
Updated on: 24 Apr 2021, 13:32
Originally posted by TarunKumar1234 on 24 Apr 2021, 13:29.
Last edited by TarunKumar1234 on 24 Apr 2021, 13:32, edited 1 time in total.
Last edited by TarunKumar1234 on 24 Apr 2021, 13:32, edited 1 time in total.
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Set 1: 3M and 1W
P (selecting 2 men) = \(\frac{3C2}{4C2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Set 2: 1M and 2W
P (selecting 2 women) = \(\frac{2C2}{3C2}\) = \(\frac{1}{3}\)
So, P (both events) = P (selecting 2 men)* P (selecting 2 women) =\(\frac{ 1}{2}\) * \(\frac{1}{3}\) = \(\frac{1}{6}\)
I think B.
P (selecting 2 men) = \(\frac{3C2}{4C2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Set 2: 1M and 2W
P (selecting 2 women) = \(\frac{2C2}{3C2}\) = \(\frac{1}{3}\)
So, P (both events) = P (selecting 2 men)* P (selecting 2 women) =\(\frac{ 1}{2}\) * \(\frac{1}{3}\) = \(\frac{1}{6}\)
I think B.
I don't know how I made this mistake... 
