Respuesta :
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
[tex]P(A)=\frac{1}{5}[/tex] and [tex]P(A')=1-P(A)=1-\frac{1}{5}=\frac{4}{5}[/tex]
[tex]P(B)=\frac{3}{8}[/tex] and [tex]P(B')=\frac{5}{8}[/tex]
[tex]P(C)=\frac{2}{7}[/tex] and [tex]P(C')=\frac{5}{7}[/tex]
We need to find the probability that exactly two of the three players will win but one will lose.
[tex]Probability=P(A\cap B\cap C')+P(A\cap B'\cap C)+P(A'\cap B\cap C)[/tex]
[tex]Probability=P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)[/tex]
[tex]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}[/tex]
[tex]Probability=\frac{3}{56}+\frac{1}{28}+\frac{3}{35}[/tex]
[tex]Probability=\frac{7}{40}[/tex]
The probability that exactly two of the three players will win but one will lose is [tex]\frac{7}{40}[/tex].
Therefore the correct option is E.
