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Eight women and two men are available to serve on a committe
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05 Aug 2018, 14:59
05 Aug 2018, 14:59
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Question Stats:
75% (01:49) correct
25% (01:55) wrong based on 80 sessions
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Eight women and two men are available to serve on a committee. If three people are picked, what is the probability that the committee includes at least one man?
(A) \(\frac{1}{32}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{2}{5}\)
(D) \(\frac{7}{15}\)
(E) \(\frac{8}{15}\)
(A) \(\frac{1}{32}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{2}{5}\)
(D) \(\frac{7}{15}\)
(E) \(\frac{8}{15}\)
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Joined: 06 Aug 2018
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Re: Eight women and two men are available to serve on a committe
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09 Aug 2018, 01:39
09 Aug 2018, 01:39
We can use the rule of probability:
The sum of the probabilities of all possible outcomes is 1( ex: p(a)+p)(not a) =1)
P(picking NO MAN) + P(picking 1 MAN) + P(picking 2 MAN) = 1
The probability of at least one man = P(1 MAN) + P(2 MAN)= 1- P(NO MAN)
P(NO MAN)= P( ALL WOMAN) = 8/10 * 7/9 * 6/8
= 7/15
The probability of at least one man = 1- P(NO MAN) = 1-7/15
=8/15.
Answer choice (E)
The sum of the probabilities of all possible outcomes is 1( ex: p(a)+p)(not a) =1)
P(picking NO MAN) + P(picking 1 MAN) + P(picking 2 MAN) = 1
The probability of at least one man = P(1 MAN) + P(2 MAN)= 1- P(NO MAN)
P(NO MAN)= P( ALL WOMAN) = 8/10 * 7/9 * 6/8
= 7/15
The probability of at least one man = 1- P(NO MAN) = 1-7/15
=8/15.
Answer choice (E)
Re: Eight women and two men are available to serve on a committe
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23 Aug 2018, 07:47
23 Aug 2018, 07:47
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Explanation
Because this is an “at least” question, use the 1 – x shortcut:
(The probability of picking at least one man) + (The probability of picking no men) = 1
The probability of picking no men is an and setup: woman and woman and woman.
For the first choice, there are 8 women out of 10 people: \(\frac{8}{10}=\frac{4}{5}\).
For the second choice, there are \(\frac{7}{9}\)(because one woman has already been chosen).
For the third choice, there are \(\frac{6}{8}=\frac{3}{4}\).
Multiply the three probabilities together to find the probability that the committee will be comprised of woman and woman and woman:
\(\frac{4}{5} \times \frac{7}{9} \times \frac{3}{4}=\frac{7}{15}\).
To determine the probability of picking at least one man, subtract this result from 1:
\(1- \frac{7}{15}=\frac{8}{15}\).
Because this is an “at least” question, use the 1 – x shortcut:
(The probability of picking at least one man) + (The probability of picking no men) = 1
The probability of picking no men is an and setup: woman and woman and woman.
For the first choice, there are 8 women out of 10 people: \(\frac{8}{10}=\frac{4}{5}\).
For the second choice, there are \(\frac{7}{9}\)(because one woman has already been chosen).
For the third choice, there are \(\frac{6}{8}=\frac{3}{4}\).
Multiply the three probabilities together to find the probability that the committee will be comprised of woman and woman and woman:
\(\frac{4}{5} \times \frac{7}{9} \times \frac{3}{4}=\frac{7}{15}\).
To determine the probability of picking at least one man, subtract this result from 1:
\(1- \frac{7}{15}=\frac{8}{15}\).
