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

1

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

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

= \(\frac{6}{2N}\)

= \(\frac{3}{N}\)

Answer: C

Cheers, Brent

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