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

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

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New post 12 Jul 2015, 08:07
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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
[Reveal] Spoiler: OA

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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 12 Jul 2015, 08:13
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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.
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Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 12 Jul 2015, 08:22
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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
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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 12 Jul 2015, 08:36
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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.
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Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
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Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 12 Jul 2015, 08:52
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Expert's post
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,
\(\frac{7C1*11C3}{18C4} = 7*\frac{11!}{8!3!}*\frac{4!14!}{18!}=7*\frac{11*10*9}{3!}*\frac{4!}{18*17*16*15}\)........

\(\frac{7*4*11*10*9}{18*17*16*15}\)
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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 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).
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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 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?

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Re: What is the probability that a 4 person committee chosen at random fro [#permalink]

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New post 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.
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Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

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

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New post 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!

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What is the probability that a 4 person committee chosen at random fro   [#permalink] 28 Sep 2017, 13:30
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