Question

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

Answer 1

Answer:

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

Answer 2

Answer:

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