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Alice, Benjamin, and Carol each try independently...

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Alice, Benjamin, and Carol each try independently...

by BTGmoderatorLU » Fri Jan 05, 2018 7:46 am
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

The OA is E.

I'm confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
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by ErikaPrepScholar » Fri Jan 05, 2018 8:09 am
There are three possibilities for two players to win but one player to lose:

A and B win, C loses
B and C win, A loses
A and C win, B loses

So we will add together the probabilities of all three possibilities to get the total probability:

A and B win, C loses
$$\frac{1}{5}\left(A\ wins\right)\cdot\frac{3}{8}\left(B\ wins\right)\cdot\left(1-\frac{2}{7}\right)\left(C\ loses\right)$$ $$\frac{1}{5}\cdot\frac{3}{8}\cdot\frac{5}{7}$$ $$\frac{15}{280}$$

B and C win, A loses
$$\frac{3}{8}\left(B\ wins\right)\cdot\frac{2}{7}\left(C\ wins\right)\cdot\left(1-\frac{1}{5}\right)\left(A\ loses\right)$$ $$\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{4}{5}$$ $$\frac{24}{280}$$

A and C win, B loses
$$\frac{1}{5}\left(A\ wins\right)\cdot\frac{2}{7}\left(C\ wins\right)\cdot\left(1-\frac{3}{8}\right)\left(B\ loses\right)$$ $$\frac{1}{5}\cdot\frac{2}{7}\cdot\frac{5}{8}$$ $$\frac{10}{280}$$

Adding those together gives

$$\frac{15}{280}+\frac{24}{280}+\frac{10}{280}=\frac{49}{280}=\frac{7}{40}$$
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