It is currently 28 May 2020, 05:01

### 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).

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
GRE Instructor
Joined: 10 Apr 2015
Posts: 3253
Followers: 124

Kudos [?]: 3642 [0], given: 61

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

Question Stats:

73% (01:32) correct 26% (01:41) wrong based on 15 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
Sign up for my free GRE Question of the Day emails

GRE Instructor
Joined: 10 Apr 2015
Posts: 3253
Followers: 124

Kudos [?]: 3642 [1] , given: 61

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
Sign up for my free GRE Question of the Day emails

Intern
Joined: 27 Jun 2019
Posts: 40
Followers: 0

Kudos [?]: 8 [1] , given: 3

Re: Joe is in a group consisting of N people (N > 4). [#permalink]  29 Aug 2019, 06:13
1
KUDOS
Great Question. Thanks!
Re: Joe is in a group consisting of N people (N > 4).   [#permalink] 29 Aug 2019, 06:13
Display posts from previous: Sort by

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

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.