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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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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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34
[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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Bunuel 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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34


Total Members = 6+7+5 = 18


Total ways of making 4 person committee = 18C4 = 3060


Total Members without women = 6+5 = 11
Total ways of choosing e person committee without any women = (6+5)C4 = 330

favorable probbility = 1- (330/3060) = 1- (11/102) = 91/102

Answer: option D
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What is the probability that a 4-person committee chosen at random fro [#permalink]

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New post 21 Oct 2016, 14:10
Bunuel 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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34


The probability of at least one woman on the four-person team is the same as (1 - probability of no women on the team).

Thus, the total probability is \(1 - (\frac{11}{18})(\frac{10}{17})(\frac{9}{16})(\frac{8}{15}) = 1-\frac{11}{102} = \frac{91}{102}\)

The logic behind the calculation is simple:

The probability of selecting a man or child for the first position is \(\frac{(5+6)}{(5+6+7)}=\frac{11}{18}\).

Now, the key is that each time we draw another person, we reduce both the pool we are drawing from--the number of men and children-- and the total population by 1, as there is no replacement. Thus,

The probability of selecting a man or child for the second position is \(\frac{(11-1)}{(18-1)} = \frac{10}{17}\)

The third is \(\frac{(10-1)}{(17-1)} = \frac{9}{16}\), and so on
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What is the probability that a 4-person committee chosen at random fro [#permalink]

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New post 22 Oct 2016, 11:34
Bunuel 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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34


The probability of at least one woman = (1 - Probability of no women )

Probability of no women = \(\frac{11C4}{18C4}\) => \(\frac{330}{3060} = \frac{11}{102}\)

Required Probability of at least one woman = \(1 - \frac{11}{102}\)

Hence required probability \(= \frac{91}{102}\)

Answer will be (D)

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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 24 Oct 2016, 18:38
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Bunuel 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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34


We can look at this problem in terms of only 2 possible events. Either at least 1 woman will be selected or no women will be selected (but instead men and children will be selected). This means that

P(selecting at least 1 woman) + P(selecting no women) = 1

P(selecting at least 1 woman) = 1 - P(selecting no women)

Thus, if we can determine the probability that no women will be selected, we’ll quickly be able to calculate the probability that at least one woman is selected.

The group contains 18 individuals, of which 11 are not women, and we will select 4 people for a committee. The probability that no women are selected in 4 selections is:

11/18 x 10/17 x 9/16 x 8/15

(11 x 10 x 9 x 8)/(18 x 17 x 16 x 15)

(11 x 2 x 5 x 9 x 8)/(2 x 9 x 17 x 2 x 8 x 3 x 5)

11 /( 17 x 2 x 3)

11/102

Thus the probability of selecting at least one woman is:

1 - 11/102 = 91/102

Answer: D
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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 17 Dec 2017, 22:41
Bunuel 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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34


The answer is D

I found this question a little tricky .

Now we have to find the probability of no women in the committee.
so we can select the members from children and men so total number of men and children =6+5=11
Total number of persons to choose from =5+6+7=18
So the probability of not choosing any women is =11C4/18C4
=11/102
So choosing at least one female =1-11/102=91/102
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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 23 Apr 2018, 09:36
Bunuel 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 at least 1 woman?

A. 11/102
B. 77/204
C. 77/102
D. 91/102
E. 31/34


There are 6 Men, 7 Women & 5 Children
Total 6+7+5 = 18

So, total number of possible arrangements = 18C4
Number of Possible arrangements without any women = 11C4 (Selection from 6 Men and 5 Children)
Probability of 4 member committee with no women = 11C4/18C4 = 11*10*9*8/18/17/16/15 = 11*2*1*1/2/17/2/3 = 11/102

Probability of 4 member committee with atleast 1 woman = 1- (11/102) = 91/102

Answer D
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Re: What is the probability that a 4-person committee chosen at random fro   [#permalink] 23 Apr 2018, 09:36
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