• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A committee of 2 people is to be selected out of ##### This topic has expert replies Legendary Member Posts: 563 Joined: 14 Oct 2017 Followed by:3 members ### A committee of 2 people is to be selected out of by VJesus12 » Thu Mar 15, 2018 4:23 am A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher? A. 1/4 B. 1/3 C. 2/3 D. 6/7 E. 8/9 The OA is D. Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i> Legendary Member Posts: 2666 Joined: 14 Jan 2015 Location: Boston, MA Thanked: 1153 times Followed by:125 members GMAT Score:770 by DavidG@VeritasPrep » Thu Mar 15, 2018 5:37 am VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher? A. 1/4 B. 1/3 C. 2/3 D. 6/7 E. 8/9 The OA is D. Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i> Yes! (You can think of probability as a ratio of combinations or permutations.) Useful equation P(x) = 1 - P(not x) P( at least 1 preacher) = 1 - P(no preachers) P(no preachers) = # ways we can select committee with no preachers/# ways we can select committee total # ways we can select no preachers: Both selections would be teachers. If there are 3 to choose from, we want 3C2 = 3*2/2! = 3. # ways we can select committee total: There are 7 people. We want to, or 7C2 = 7*6/2! = 21 P(no preachers) = 3/21 = 1/7 1 - P(no preachers) = 1 - 1/7 = 6/7. The answer is D Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

Legendary Member
Posts: 2666
Joined: 14 Jan 2015
Location: Boston, MA
Thanked: 1153 times
Followed by:125 members
GMAT Score:770
by DavidG@VeritasPrep » Thu Mar 15, 2018 5:41 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?

A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9

The OA is D.

Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
You can also go pick by pick. We want 1 - P( no preachers)

P(no preachers) = P(pick 1 not a preacher) * P(pick 2 not a preacher given that pick 1 was not.)
P(pick 1 not a preacher) = 3/7
P(pick 2 not a preacher given that pick 1 was not) = 2/6
P(pick 1 not a preacher) * P(pick 2 not a preacher given that pick 1 was not) = 3/7 * 2/6 = 6/42 = 1/7
1 - P(no preachers) = 1 - 1/7 = 6/7, or D

The gist is that you can think of probability as a ratio of combinations, or you can go pick by pick. (Just make sure that if you go pick by pick that you consider every possible order of selection.)
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save \$100 off any live Veritas Prep GMAT Course

### GMAT/MBA Expert

GMAT Instructor
Posts: 13881
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1257 members
GMAT Score:770
by Brent@GMATPrepNow » Thu Mar 15, 2018 5:46 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?

A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
Here's an approach that uses probability rules.

We want P(select at least 1 preacher)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A NOT happening)
So, here we get: P(getting at least 1 preacher) = 1 - P(NOT getting at least 1 preacher)
What does it mean to not get at least 1 preacher? It means getting zero preachers.
So, we can write: P(getting at least 1 preacher) = 1 - P(getting zero preachers)

P(getting zero preachers)
P(getting zero preachers) = P(1st selection is teacher AND 2nd selection is teacher)
= P(1st selection is teacher) x P(2nd selection is teacher)
= 3/7 x 2/6
= 1/7

So, P(getting at least 1 preacher) = 1 - 1/7
=6/7

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Watch these video reviews of my course
And check out these free resources

Legendary Member
Posts: 860
Joined: 07 Sep 2017
Thanked: 6 times
Followed by:3 members
by Vincen » Thu Mar 15, 2018 6:45 am
Hello Vjesus12.

This is how I would solve it: I'd use combinations and then probability.

There are 7 people and we have to pick 2, this can be done from $$7C2=\frac{7!}{5!2!}=\frac{7\cdot6\cdot5!}{5!\cdot2}=\frac{42}{2}=21.$$ Now:

The number of ways to pick 1 preacher is given by: $$3C1\cdot4C1=\frac{3!}{2!1!}\cdot\frac{4!}{3!1!}=3\cdot4=12.$$ The number of ways to pick 2 preachers is given by: $$4C2=\frac{4!}{2!2!}=\frac{12}{2}=6.$$ Hence, the total of ways to pick at least 1 preacher is 12+6 = 18.

Now, the probability of pick at least 1 preacher is: $$P=\frac{18}{21}=\frac{6}{7}.$$ Therefore, the correct answer is the option D.

### GMAT/MBA Expert

GMAT Instructor
Posts: 4430
Joined: 25 Apr 2015
Location: Los Angeles, CA
Thanked: 43 times
Followed by:21 members
by Scott@TargetTestPrep » Mon Mar 19, 2018 5:45 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?

A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
To determine how many ways to select at least one preacher, we can use the formula:

Number of ways to select at least 1 preacher = total number of ways to select the committee - number of ways to select NO preachers

Total number of ways to select the committee of 2 people is:

7C2 = 7!/[2!(7-2)!] = (7 x 6)/2! = 21

The number of ways to select NO preachers in the committee is:

3C2 = (3 x 2)/2! = 3

Thus, the number of ways to select at least 1 preacher is 21 - 3 = 18.

So the probability that the committee will be composed of at least 1 preacher is 18/21 = 6/7.

Alternate Solution:

We use the rule of the complement, which says: The probability of picking AT LEAST one preacher = 1 - the probability of picking NO preachers.

Note that the probability of picking NO preachers is the same as the probability of picking all teachers for this committee of 2. Thus, the probability of picking 2 teachers is 3/7 x 2/6 = 1/7.

Thus, the probability of picking at least one preacher = 1 - 1/7= 6/7.