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# Joe is in a group consisting of N people (N > 4).

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GRE Instructor
Joined: 10 Apr 2015
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Joe is in a group consisting of N people (N > 4). [#permalink]  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) $$\frac{N^2-1}{5N}$$
[Reveal] Spoiler: OA

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Brent Hanneson – Creator of greenlighttestprep.com

GRE Instructor
Joined: 10 Apr 2015
Posts: 1743
Followers: 58

Kudos [?]: 1660 [1] , given: 8

Re: Joe is in a group consisting of N people (N > 4). [#permalink]  16 May 2019, 05:14
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Expert's post
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) $$\frac{N^2-1}{5N}$$

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 $$\frac{6}{(N)(N-1)(N-2)}$$

= $$\frac{6(N-1)(N-2)}{2(N)(N-1)(N-2)}$$

= $$\frac{6}{2N}$$

= $$\frac{3}{N}$$

Cheers,
Brent

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Brent Hanneson – Creator of greenlighttestprep.com

Re: Joe is in a group consisting of N people (N > 4).   [#permalink] 16 May 2019, 05:14
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