Conditional probability of C given two independent events

Question

Suppose that a machine depends on the working state of two components A and B.

1. If both $A$ and $B$ do not function then the probability (say $C$) of the machine to work is $0.3$
2. If both $A$ and $B$ work then $P(C) = 0.9$
3. If exactly one of $A$ or $B$ works then $P(C) = 0.5$

The probability of $A$ to work is $P(A) = 0.7$, while $P(B) = 0.8$. The working state of $A$ does not affect the state of $B$ and vice versa (I suppose this means that $A$ and $B$ are independent, right?). How can I calculate the probability of the machine to work or $P(C)$ ? And, additionally if the machine works, what is the possibility of both $A$ and $B$ work?

I suppose that 1-3 are the results of $P(C|A^{c} \cap B^{c}), P(C|A \cap B)$ and $P(C|A^{c}B \cup AB^{c})$ and $P(A \cap B)=P(A)P(B), P(A \cup B) = P(A)+P(B)-P(A \cap B)$ but still I cannot figure how to compute P(C) out of Bayes rule... For the latter, I understand that I have to compute $P(A \cap B|C)$, but still I need $P(C)$. What am I missing? It's been two days that I am writing around equations but still nothing! Thanks in advance!

Answer 1

The three events $A\cap B, A^c\cap B^c$, and $(A^c\cap B)\cup (A\cap B^c)$ are a partition of the sample space, and thus you can use the law of total probability to get $$\begin{align} P(C) &= P(C\mid A\cap B)P(A\cap B)\\ & +P(C\mid A^c\cap B^c)P(A^c\cap B^c)\\ & + P(C\mid (A^c\cap B)\cup (A\cap B^c))P((A^c\cap B)\cup (A\cap B^c)). \end{align}$$ You are given the three conditional probabilities on the right of this equation, and you know how to compute $P(A\cap B)$ from the knowledge (or assumption) that $A$ and $B$ are independent. I will only hint that $A^c$ and $B^c$ are independent, as are $A$ and $B^c$, and $A^c$ and $B$ independent events. Finally, $P(A^c\cap B)\cup (A\cap B^c)) = P(A^c\cap B)+ P(A\cap B^c)$. Can you tell why?

Answer 2

I use $B^c$ to denote the event that $B$ does not work. \begin{align*} P(C) &= P(C \cap A \cap B) + P(C \cap A \cap B^c) + P(C \cap A^c \cap B) + P(C \cap A^c \cap B^c). \end{align*}

For each of the terms, you can rewrite in terms of the conditional probabilities. E.g., $$P(C \cap A \cap B) = P(C \mid A \cap B) \cdot P(A \cap B)=P(C \mid A \cap B) \cdot P(A)\cdot P(B)$$