GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Aug 2018, 09:40

### 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

# What is the probability that a 4 person committee chosen at random fro

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 29 Apr 2015
Posts: 862
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

12 Jul 2015, 08:07
3
00:00

Difficulty:

55% (hard)

Question Stats:

73% (02:37) correct 27% (03:02) wrong based on 130 sessions

### HideShow timer Statistics

What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

Current Student
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

12 Jul 2015, 08:13
1
reto wrote:
What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

Total possible selections = 4 out of 18 group members = 18C4

Favorable selections = 1 out of 7 women and 3 out of 11 (= 6 men + 5 children) = 7C1 * 11C3

Thus the required probability = 7C1*11C3 / 18C4 = 77/204. Thus A is the correct answer.
Retired Moderator
Joined: 29 Apr 2015
Posts: 862
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

12 Jul 2015, 08:22
1
Engr2012 wrote:
reto wrote:
What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

Total possible selections = 4 out of 18 group members = 18C4

Favorable selections = 1 out of 7 women and 3 out of 11 (= 6 men + 5 children) = 7C1 * 11C3

Thus the required probability = 7C1*11C3 / 18C4 = 77/204. Thus A is the correct answer.

Thanks, thats a nice aproach too!

Could you elaborate on how you get to the final answer of 77/204 (from here: 7C1*11C3 / 18C4 = 77/204)?

Official Solution: 4 * 11/18 * 7/17 * 10/16 *9/15
_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

Current Student
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

12 Jul 2015, 08:36
1
reto wrote:
Engr2012 wrote:
reto wrote:
What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

Total possible selections = 4 out of 18 group members = 18C4

Favorable selections = 1 out of 7 women and 3 out of 11 (= 6 men + 5 children) = 7C1 * 11C3

Thus the required probability = 7C1*11C3 / 18C4 = 77/204. Thus A is the correct answer.

Thanks, thats a nice aproach too!

Could you elaborate on how you get to the final answer of 77/204 (from here: 7C1*11C3 / 18C4 = 77/204)?

Official Solution: 4 * 11/18 * 7/17 * 10/16 *9/15

Sure.

18C4 = 18! / (4!*14!) .... as nCr = n!/(r!*(n-r)!) = 17*15*4*3

*and the numerator = 7C1*11C3 = 7*11*5*3

Thus, the final answer = (7*11*5*3) / (17*15*4*3) = 77/204.
Math Expert
Joined: 02 Aug 2009
Posts: 6559
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

12 Jul 2015, 08:52
1
reto wrote:
Engr2012 wrote:
reto wrote:
What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

Total possible selections = 4 out of 18 group members = 18C4

Favorable selections = 1 out of 7 women and 3 out of 11 (= 6 men + 5 children) = 7C1 * 11C3

Thus the required probability = 7C1*11C3 / 18C4 = 77/204. Thus A is the correct answer.

Thanks, thats a nice aproach too!

Could you elaborate on how you get to the final answer of 77/204 (from here: 7C1*11C3 / 18C4 = 77/204)?

Official Solution: 4 * 11/18 * 7/17 * 10/16 *9/15

hi,
$$\frac{7C1*11C3}{18C4} = 7*\frac{11!}{8!3!}*\frac{4!14!}{18!}=7*\frac{11*10*9}{3!}*\frac{4!}{18*17*16*15}$$........

$$\frac{7*4*11*10*9}{18*17*16*15}$$
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Senior Manager
Joined: 28 Jun 2015
Posts: 295
Concentration: Finance
GPA: 3.5
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

12 Jul 2015, 10:09
Sample space = 18C4 = (18*17*16*14)/(1*2*3*4) = 3060.

Favourable events = (7C1) * (11C3) = 1155.

Probability = 1155/3060 = 231/612 = 77/204. Ans (A).
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Current Student
Joined: 22 Apr 2015
Posts: 48
Location: United States
GMAT 1: 620 Q46 V27
GPA: 3.86
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

15 Jul 2015, 15:07
Can someone explain the reasoning behind 11C3 is it to choose the remaining members from the men and children. If so, why is that important if we just want to determine the probability of choosing exactly 1 woman from the group?
Current Student
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

15 Jul 2015, 15:16
xLUCAJx wrote:
Can someone explain the reasoning behind 11C3 is it to choose the remaining members from the men and children. If so, why is that important if we just want to determine the probability of choosing exactly 1 woman from the group?

Yes, you are correct that the logic behind 11C3 is to choose 3 out of the remaining 11 'people'.

You need to know this number to calculate the number of ways you can select 'remaining' 3 people to make up the 4 member committee. When you just take the probability of selecting 1 women , you are still missing the probability of remaining 3 people. Total probability will be affected by both, probability of selecting 1 out of 7 women and probability of selecting 3 out of remaining 11 men+children.

Hope this helps.
Intern
Joined: 12 Jan 2017
Posts: 3
What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

28 Sep 2017, 13:30
Here is how I solved the problem:

Step 1) Write out all of the possibilities for the four slots
(W= Women, NW= Not a Women).
- W, NW, NW, NW
- NW, W, NW, NW
- NW, NW, W, NW
- NW, NW, NW, W

There are 4 different arrangements.

Step 2) Calculate the probability of each arrangement separately combining them with "and" (meaning multiply). And remember that with each selection, the amount available becomes one smaller.

Chance of W, NW, NW, NW = (7/18)(11/17)(10/16)(9/15)
Chance of NW, W, NW, NW = (11/18)(7/17)(10/16)(9/15)
Now stop.. Realize that no matter the arrangement, the numerators and denominators will remain the same.

Step 3) Simplify and calculate 4*[(7/18)(11/17)(10/16)(9/15)] = 77/204

*With practice you should be able to quickly determine that there are four ways of arrangement and that they will all yield the same results, which puts you straight to step 3!
Intern
Joined: 12 Sep 2017
Posts: 28
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

05 Jun 2018, 12:47
reto wrote:
What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

Hi

Can someone please tell why are we not calculating probability separately for Men and Children instead of taking them together? The way i tried was,
7C1x6C3 + 7C1X6C2X5C1 + 7C1X6C1X5C2 but i am not able to arrive at the answer.

Thanks
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3690
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

06 Jun 2018, 01:12
1
@s wrote:
Hi

Can someone please tell why are we not calculating probability separately for Men and Children instead of taking them together? The way i tried was,
7C1x6C3 + 7C1X6C2X5C1 + 7C1X6C1X5C2 but i am not able to arrive at the answer.

Thanks

Hey @s ,

You need to understand that the condition is given only on the women. So, whenever that is the case, you should not put conditions on the remaining categories.

Here, I need to make sure that out of 4 people, I must have exactly one women. This means I will choose one women out of the 7 and then choose 3 more people out of the remaining 11 (No matter whether all are males or children).

Does that make sense?
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.

Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free

Intern
Joined: 12 Sep 2017
Posts: 28
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

07 Jun 2018, 03:26
abhimahna wrote:
@s wrote:
Hi

Can someone please tell why are we not calculating probability separately for Men and Children instead of taking them together? The way i tried was,
7C1x6C3 + 7C1X6C2X5C1 + 7C1X6C1X5C2 but i am not able to arrive at the answer.

Thanks

Hey @s ,

You need to understand that the condition is given only on the women. So, whenever that is the case, you should not put conditions on the remaining categories.

Here, I need to make sure that out of 4 people, I must have exactly one women. This means I will choose one women out of the 7 and then choose 3 more people out of the remaining 11 (No matter whether all are males or children).

Does that make sense?

Yes. Thank you so much.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3183
Location: United States (CA)
Re: What is the probability that a 4 person committee chosen at random fro  [#permalink]

### Show Tags

07 Jun 2018, 17:06
reto wrote:
What is the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman?

A. 77/204
B. 77/832
C. 11/77
D. 308/1411
E. 22/832

The number of ways to select one woman is 7C1 = 7.

The number of ways to select 3 men and/or children is 11C3 = (11 x 10 x 9)/3! = (11 x 10 x 9)/(3 x 2) = 11 x 5 x 3

The number of ways to select 4 people from 18 is 18C4:

18C4 = (18 x 17 x 16 x 15)/4! = (18 x 17 x 16 x 15)/(4 x 3 x 2) = 3 x 17 x 4 x 15

So the probability is:

(7 x 11 x 5 x 3)/(3 x 17 x 4 x 15) = (7 x 11)/(3 x 17 x 4) = 77/204

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: What is the probability that a 4 person committee chosen at random fro &nbs [#permalink] 07 Jun 2018, 17:06
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.