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Joined: 10 Apr 2015
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Joe is in a group consisting of N people (N > 4).
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15 May 2019, 08:47
15 May 2019, 08:47
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Joe is in a group consisting of N people (N > 4).
If 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?
A) \(\frac{N^2-2N-6}{(N-2)^2+6}\)
B) \(\frac{2N+3}{N^2-N}\)
C) \(\frac{3}{N}\)
D) \(\frac{N}{(N-1)(N-2)}\)
E) \(\frac{N^2-1}{5N}\)
If 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?
A) \(\frac{N^2-2N-6}{(N-2)^2+6}\)
B) \(\frac{2N+3}{N^2-N}\)
C) \(\frac{3}{N}\)
D) \(\frac{N}{(N-1)(N-2)}\)
E) \(\frac{N^2-1}{5N}\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Joe is in a group consisting of N people (N > 4).
[#permalink]
16 May 2019, 05:14
16 May 2019, 05:14
2
GreenlightTestPrep wrote:
Joe is in a group consisting of N people (N > 4). If 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?
A) \(\frac{N^2-2N-6}{(N-2)^2+6}\)
B) \(\frac{2N+3}{N^2-N}\)
C) \(\frac{3}{N}\)
D) \(\frac{N}{(N-1)(N-2)}\)
E) \(\frac{N^2-1}{5N}\)
A) \(\frac{N^2-2N-6}{(N-2)^2+6}\)
B) \(\frac{2N+3}{N^2-N}\)
C) \(\frac{3}{N}\)
D) \(\frac{N}{(N-1)(N-2)}\)
E) \(\frac{N^2-1}{5N}\)
My solution:
If there are N people, then we can select 3 people in NC3 (N choose 3) different ways.
NC3 = \(\frac{(N)(N-1)(N-2)}{3!}\) = \(\frac{(N)(N-1)(N-2)}{6}\)
So, this is our denominator
In how many of the (N)(N-1)(N-2)/6 possible outcomes is Joe chosen?
Well, once we make Joe one of the selected people, there are N-1 people remaining.
We can select 2 people from the remaining N-1 people in (N-1)C2 ways
(N-1)C2 = \(\frac{(N-1)(N-2)}{2!}\) = \(\frac{(N-1)(N-2)}{2}\)
This is our numerator
So, P(Joe is selected) = \(\frac{(N-1)(N-2)}{2}\) ÷ \(\frac{(N)(N-1)(N-2)}{6}\)
= \(\frac{(N-1)(N-2)}{2}\) x \(\frac{6}{(N)(N-1)(N-2)}\)
= \(\frac{6(N-1)(N-2)}{2(N)(N-1)(N-2)}\)
= \(\frac{6}{2N}\)
= \(\frac{3}{N}\)
Answer: C
Cheers,
Brent
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Re: Joe is in a group consisting of N people (N > 4).
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29 Aug 2019, 06:13
29 Aug 2019, 06:13
1
Great Question. Thanks!
