It is currently 21 Jul 2019, 09:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:
GRE Instructor
Joined: 10 Apr 2015
Posts: 2175
Followers: 64

Kudos [?]: 1986 [0], given: 20

Joe is in a group consisting of N people (N > 4). [#permalink]  15 May 2019, 08:47
Expert's post
00:00

Question Stats:

75% (01:26) correct 25% (00:00) wrong based on 4 sessions
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

_________________

Brent Hanneson – Creator of greenlighttestprep.com

GRE Instructor
Joined: 10 Apr 2015
Posts: 2175
Followers: 64

Kudos [?]: 1986 [1] , given: 20

Re: Joe is in a group consisting of N people (N > 4). [#permalink]  16 May 2019, 05:14
1
KUDOS
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

RELATED VIDEO FROM MY COURSE

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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