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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 tagged by: VJesus12 ##### This topic has 4 expert replies and 1 member reply ### Top Member ## A committee of 2 people is to be selected out of 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. Legendary Member Joined 07 Sep 2017 Posted: 831 messages Followed by: 3 members Upvotes: 6 Top Reply 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 Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 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. 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

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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.
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.)

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Save $100 off any live Veritas Prep GMAT Course Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now! ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12987 messages Followed by: 1249 members Upvotes: 5254 GMAT Score: 770 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 â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2801 messages Followed by: 18 members Upvotes: 43 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 scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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