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To choose a committee, the names of all pos- sible members a
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07 Jan 2021, 09:45
07 Jan 2021, 09:45
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To choose a committee, the names of all possible members are written on identical cards, and five cards are drawn at random. If there are 14 men and 16 women among the possible members, what is the probability that all committee members are women?
A. \(\frac{1}{2}\)
B. \(\frac{1}{32}\)
C. \(\frac{8}{15}\)
D. \((\frac{8}{15})^5\)
E. \(\frac{8}{261}\)
Kudos for the right answer and explanation
A. \(\frac{1}{2}\)
B. \(\frac{1}{32}\)
C. \(\frac{8}{15}\)
D. \((\frac{8}{15})^5\)
E. \(\frac{8}{261}\)
Kudos for the right answer and explanation
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Re: To choose a committee, the names of all pos- sible members a
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08 Jan 2021, 07:06
08 Jan 2021, 07:06
1
Carcass wrote:
To choose a committee, the names of all possible members are written on identical cards, and five cards are drawn at random. If there are 14 men and 16 women among the possible members, what is the probability that all committee members are women?
A. \(\frac{1}{2}\)
B. \(\frac{1}{32}\)
C. \(\frac{8}{15}\)
D. \((\frac{8}{15})^5\)
E. \(\frac{8}{261}\)
Kudos for the right answer and explanation
A. \(\frac{1}{2}\)
B. \(\frac{1}{32}\)
C. \(\frac{8}{15}\)
D. \((\frac{8}{15})^5\)
E. \(\frac{8}{261}\)
Kudos for the right answer and explanation
P(all women) = P(1st selection is woman AND 2nd selection is woman AND 3rd is woman AND 4th is woman AND 5th is woman)
= P(1st selection is woman) x P(2nd selection is woman) x P(3rd is woman) x P(4th is woman) x P(5th is woman)
\(= \frac{16}{30} \times \frac{15}{29} \times \frac{14}{28} \times \frac{13}{27} \times \frac{12}{26}\)
TIME-SAVER TIP: Notice that the second fraction, 15/29, has a prime number (29) in the denominator. Since none of the numerators are divisible by 29, we can conclude that the resulting product must have a denominator that's divisible by 29.
Since the denominators in answer choices A, B, C and D are definitely not divisible by 29, the correct answer is E.
Or you could just complete the calculations, in which case you'll find the resulting product is \(\frac{8}{261}\)
gmatclubot
Re: To choose a committee, the names of all pos- sible members a [#permalink]
08 Jan 2021, 07:06
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