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09 Aug 2025, 12:47
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D
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Difficulty:
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56% (01:46) correct
44%
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A committee of three people is to be randomly selected from a group of people. How many people are in the group?
(1) The probability that two members of the group, Bruce and Eileen, will be among those selected for the committee is 1/15.
(2) The number of different committees that could be chosen from the group is 120.
(1) The probability that two members of the group, Bruce and Eileen, will be among those selected for the committee is 1/15.
(2) The number of different committees that could be chosen from the group is 120.
Updated on: 09 Aug 2025, 13:38
Originally posted by gmatophobia on 09 Aug 2025, 13:37.
Last edited by gmatophobia on 09 Aug 2025, 13:38, edited 1 time in total.
Last edited by gmatophobia on 09 Aug 2025, 13:38, edited 1 time in total.
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MartyMurray
Assume that there are \(n\) members in the group.
Statement 1
(1) The probability that two members of the group, Bruce and Eileen, will be among those selected for the committee is 1/15.
From a group of \(n\) members, the number of possible committees of 3 is \(^nC_3\)
\(^nC_3 = \frac{n*(n-1)*(n-2)}{3*2}\)
The number of committees that include both Bruce and Eileen is the number of ways to choose the third member from the remaining \(n -2\) members. This can be done in \(n - 2\) ways.
Required Probability = \(\frac{(n-2)}{ \frac{n*(n-1)*(n-2)}{3*2}}\)
\(\frac{(n-2)}{ \frac{n*(n-1)*(n-2)}{3*2}} = \frac{1}{15}\)
\(\frac{6}{n*(n-1)} = \frac{1}{15}\)
\(n(n-1) = 90\)
n = 10
So from statement (1), we can conclude there are 10 people.
Sufficient.
Statement 2
(2) The number of different committees that could be chosen from the group is 120.
From a group of \(n\) members, the number of possible committees of 3 is \(^nC_3\)
\(^nC_3 = \frac{n*(n-1)*(n-2)}{3*2}\)
\(\frac{n*(n-1)*(n-2)}{6} = 120\)
\(n*(n-1)*(n-2) = 720\)
\(n = 10\)
So from statement (2) as well, we can conclude there are 10 people.
Sufficient.
Option D
