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

Answer:

The correct option is E.

Step-by-step 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

$P(A)=\frac{1}{5}$ and $P(A')=1-P(A)=1-\frac{1}{5}=\frac{4}{5}$

$P(B)=\frac{3}{8}$ and $P(B')=\frac{5}{8}$

$P(C)=\frac{2}{7}$ and $P(C')=\frac{5}{7}$

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

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

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

$Probability=\frac{1}{5}\cdot\frac{3}{8}\cdot\frac{5}{7}+\frac{1}{5}\cdot\frac{5}{8}\cdot\frac{2}{7}+\frac{4}{5}\cdot\frac{3}{8}\cdot\frac{2}{7}$

$Probability=\frac{3}{56}+\frac{1}{28}+\frac{3}{35}$

$Probability=\frac{7}{40}$

The probability that exactly two of the three players will win but one will lose is $\frac{7}{40}$.

Therefore the correct option is E.