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

GMAT assassins aren't born, they're made,
Rich

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

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

We must individually consider each possible outcome of having two winners and one loser.

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