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A committee of three will be selected from eight employees, including

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fdfd97
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A committee of three will be selected from eight employees, including [#permalink] New post   Updated on: 26 Apr 2023, 06:38
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A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob?

A. 1/56
B. 1/28
C. 1/8
D. 15/56
E. 15/28
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D

Originally posted by fdfd97 on 26 Apr 2023, 06:25.
Last edited by Bunuel on 26 Apr 2023, 06:38, edited 1 time in total.
Renamed the topic, added the options and the OA.
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Re: A committee of three will be selected from eight employees, including [#permalink] New post   26 Apr 2023, 11:48
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fdfd97 wrote:
A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob?

A. 1/56
B. 1/28
C. 1/8
D. 15/56
E. 15/28


The number of ways of selecting three members from eight employees = \(^8C_3\) ways

As Alice is already a member, the remaining two places can be filled by 6 employees = \(^6C_2\) ways

Required Probability = \(\frac{^6C_2}{^8C_3} = \frac{15}{56}\)

Option D
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Re: A committee of three will be selected from eight employees, including [#permalink] New post   26 Apr 2023, 12:24
gmatophobia wrote:
fdfd97 wrote:
A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob?

A. 1/56
B. 1/28
C. 1/8
D. 15/56
E. 15/28


The number of ways of selecting three members from eight employees = \(^8C_3\) ways

As Alice is already a member, the remaining two places can be filled by 6 employees = \(^6C_2\) ways

Required Probability = \(\frac{^6C_2}{^8C_3} = \frac{15}{56}\)

Option D


Thanks, I was trying to calculate some confusing round about probability of selecting Alice, and got 5/56 (1/8 for Alice * 6/7*5/6) (6/7 and 5/6 to remove Bob from the count), and much later realised that I should've done it 3 more times, coz the first one is just for Alice at the first spot.

Anyhow, now I got the much simpler solution. Thanks.
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Re: A committee of three will be selected from eight employees, including [#permalink] New post   26 Apr 2023, 19:45
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fdfd97 wrote:
A committee of three will be selected from eight employees, including Alice and Bob. What is the probability that the chosen committee of three includes Alice and not Bob?

A. 1/56
B. 1/28
C. 1/8
D. 15/56
E. 15/28


The number of ways to select a committee of three, including Alice but not Bob, from eight people is 6C2 = (6 x 5)/2! = 15.

The total number of ways to select the committee is 8C3 = (8 x 7 x 6)/3! = 56

Thus, the probability that the chosen committee of three includes Alice and not Bob is 15/56 =

Answer: D
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Re: A committee of three will be selected from eight employees, including [#permalink]
26 Apr 2023, 19:45
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