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

E. 7/40

Step-by-step explanation:

(1/5) (3/8) (5/7) + (1/5) (5/8) (2/7) +

(4/5) (3/8) (2/7)

= (15+10+24) / 280

= 49/280

= 7/40

E. 7/40

Step-by-step explanation:

The following probabilities are given:

P (Alice Wins) = 1/5

P (Benj. Wins) = 3/8

P (Carol Wins) = 2/7

We can deduce the probabilities for losses:

P (Alice Loses) = 1 - P (Alice Wins) = 1 - 1/5 = 4/5

P (Benj. Loses) = 1 - P (Benj. Wins) = 1 - 3/8 = 5/8

P (Carol Loses) = 1 - P (Carol Wins) = 1 - 2/7 = 5/7

The possible outcomes that two players win and one player loses are as follows:

Alice Wins, Benj Wins, Carol Loses

Alice Loses, Benj Wins, Carol Wins

Alice Wins, Benj Loses, Carol Wins

We can compute the probabilities of each of the 3 outcomes above:

P (Alice Wins, Benj Wins, Carol Loses) = (1/5) x (3/8) x (5/7) = 3/56

Alice Loses, Benj Wins, Carol Wins = (4/5) x (3/8) x (2/7) = 3/35

Alice Wins, Benj Loses, Carol Wins = (1/5) x (5/8) x (2/7) = 2/56

P (2 wins and 1 loss)

= 3/56 + 3/35 + 2/56

= 343 / 1960

= 7/40