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. 3/140B. 1/28C. 3/56D. 3/35E. 7/40

Answer 1

The correct option is E.

Explanation:

It is given that Alice, Benjamin, and Carol each try independently to win a carnival game.

Let A, B and C represent the following events.

A = Alice win

B = Benjamin win

C = Carol win

`P(A)=(1)/(5)` and
`P(A')=1-P(A)=1-(1)/(5)=(4)/(5)`

`P(B)=(3)/(8)` and
`P(B')=(5)/(8)`

`P(C)=(2)/(7)` and
`P(C')=(5)/(7)`

We need to find the probability that exactly two of the three players will win but one will lose.

`Probability=P(A\cap B\cap C')+P(A\cap B'\cap C)+P(A'\cap B\cap C)`

`Probability=P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)`

`Probability=(1)/(5)\cdot(3)/(8)\cdot(5)/(7)+(1)/(5)\cdot(5)/(8)\cdot(2)/(7)+(4)/(5)\cdot(3)/(8)\cdot(2)/(7)`

`Probability=(3)/(56)+(1)/(28)+(3)/(35)`

`Probability=(7)/(40)`

The probability that exactly two of the three players will win but one will lose is
`(7)/(40)`.

Therefore the correct option is E.