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**Soroban** Hello, zorro!

$\displaystyle \begin{array}{ccccccccc}P(A \cap B \cap C) &=& (0.95)(0.9)(0.8) &=& 0.684 \\

P(A \cap B \cap C') &=& (0.95)(0.9)(0.2) &=& 0.171 \\

P(A \cap B' \cap C) &=& (0.85)(0.1)(0.8) &=& 0.076 \\

P(A' \cap B \cap C) &=& (0.05)(0.9)(0.8) &=& 0.036 \\ \hline

&& \text{Total:} && 0.967\end{array}$

$\displaystyle P(\text{System works}) \:=\:0.967$

$\displaystyle P(\text{Exactly 2 work }|\text{ System works}) \;=\;\frac{P(\text{Exactly 2 work }\wedge\text{ System works})}{P(\text{System works})} $

Numerator: .Exactly 2 components work:

. . $\displaystyle P(A \cap B \cap C\:\!') + P(A \cap B' \cap C) + P(A' \cap B \cap C) \:=\:0.171 + 0.076 + 0.036 \:=\:0.283 $

Denominator, from (a): .$\displaystyle P(\text{System works}) \:=\:0.967$

Therefore: .$\displaystyle P(\text{Exactly 2 work }|\text{ System works}) \;=\;\frac{0.283}{0.967} \;\approx\;0.293$

I *hope* this one is correct . . .

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