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Alice, Benjamin, and Carol each try independently to win a carnival ga
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25 Oct 2021, 05:41
25 Oct 2021, 05:41
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Question Stats:
75% (01:39) correct
25% (02:38) wrong based on 4 sessions
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Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40
A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40
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Re: Alice, Benjamin, and Carol each try independently to win a carnival ga
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25 Oct 2021, 06:02
25 Oct 2021, 06:02
1
Carcass wrote:
Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40
A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40
P(exactly 2 win) = P(A wins and B wins and C loses OR B wins and C wins and A loses OR A wins and C wins and B loses)
= P(A wins and B wins and C loses) + P(B wins and C wins and A loses) + P(A wins and C wins and B loses)
Let's calculate each probability
P(A wins and B wins and C loses) = P(A wins) x P(B wins) x P(C loses)
= 1/5 x 3/8 x 5/7
= 15/280
P(B wins and C wins and A loses) = P(B wins) x P(C wins) x P(A loses)
= 3/8 x 2/7 x 4/5
= 24/280
P(A wins and C wins and B loses) = P(A wins) x P(C wins) x P(B loses)
= 1/5 x 2/7 x 5/8
= 10/280
So, P(exactly 2 win) = 15/280 + 24/280 + 10/280
= 49/280
= 7/40
Answer: E
Cheers,
Brent
gmatclubot
Re: Alice, Benjamin, and Carol each try independently to win a carnival ga [#permalink]
25 Oct 2021, 06:02
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