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
OAE
Hi Experts ,
Please explain.
Many thanks in advance.
SJ
carnival game
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Hi jain2016,
Since we have the probabilities of each person winning a game, we can 'map out' the 3 situations in which 2 of them win and 1 of them loses:
A wins, B wins, C loses = (1/5)(3/8)(5/7) = 15/240
A wins, B loses, C wins = (1/5)(5/8)(2/7) = 10/240
A loses, B wins, C wins = (4/5)(3/8)(2/7) = 24/240
Total probability = 15/280 + 10/280 + 24/280 = 49/280 = 7/40
Final Answer: E
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Rich
Since we have the probabilities of each person winning a game, we can 'map out' the 3 situations in which 2 of them win and 1 of them loses:
A wins, B wins, C loses = (1/5)(3/8)(5/7) = 15/240
A wins, B loses, C wins = (1/5)(5/8)(2/7) = 10/240
A loses, B wins, C wins = (4/5)(3/8)(2/7) = 24/240
Total probability = 15/280 + 10/280 + 24/280 = 49/280 = 7/40
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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Hi jain2016,
The prompt asked for the probability that EXACTLY 2 of the people won, while the third person lost. To answer that question, we have to account for all the possible outcomes that 'fit' what the question asks for. I calculated each of the individual outcomes, but I still have to add them all up to get the total probability.
GMAT assassins aren't born, they're made,
Rich
The prompt asked for the probability that EXACTLY 2 of the people won, while the third person lost. To answer that question, we have to account for all the possible outcomes that 'fit' what the question asks for. I calculated each of the individual outcomes, but I still have to add them all up to get the total probability.
GMAT assassins aren't born, they're made,
Rich
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We must individually consider each possible outcome of having two winners and one loser.jain2016 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
If Alice and Benjamin win and Carol loses we have:
1/5 x 3/8 x 5/7 = 1 x 3/8 x 1/7 = 3/56
If Alice and Carol win and Benjamin loses we have:
1/5 x 5/8 x 2/7 = 1 x 1/8 x 2/7 = 2/56
If Benjamin and Carol win and Alice loses we have:
4/5 x 3/8 x 2/7 = 1/5 x 3/2 x 2/7 = 1/5 x 3 x 1/7 = 3/35
Therefore, the probability that two of them will win and one will lose is:
3/56 + 2/56 + 3/35 = 15/280 + 10/280 + 24/280 = 49/280 = 7/40
Answer: E
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