Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 563 reviewsTarget Test Prep

A committee of 2 people is to be selected out of

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Legendary Member
Posts: 2276
Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members
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>
Top
Join the discussion
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members
Location: Boston, MA
Thanked: 1153 times
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
Top
Join the discussion
User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members
Location: Boston, MA
Thanked: 1153 times
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
Top
Join the discussion

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16206
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
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

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Legendary Member
Posts: 2898
Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members
Thanked: 6 times

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.
Top
Join the discussion

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8246
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

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.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Join the discussion