(Note: the problem is translated from Japanese so there might be some mistakes)
A machine includes three devices: A1, A2, A3. This machine can have these kind of problems: Overheated (B1), Violent motion (B2) and Stop with no reason in working mode (B3). These problem can be caused only when one or some of 3 devices A1, A2, A3 be broken. The probabilities of that each devices is broken led to each problem are shown as below.
\begin{array}{|c|c|c|c|} \hline & B1 & B2 & B3 \\ \hline A1 & 0.20 & 0.30 & 0.50\\ \hline A2 & 0.40 & 0.50 & 0.10\\ \hline A3 & 0.60 & 0.10 & 0.30\\ \end{array}
Pr(A1), Pr(A2), Pr(A3) are respectively the probability that each devices are broken. Given that Pr(A3) = 2Pr(A1), Pr(A2) = 1.5Pr(A1), Pr(A1) + Pr(A2) + Pr(A3) = 1.
When the probability of B3 to occur is 0.23, find the probability that A3 is broken.
Find the probability that B1 occurs.
Find the probability that at least one of B1, B2, B3 occurs. B1, B2, B3 are independent.
For number 1, I don't know that they ask for Pr(A3) or not, but if it's Pr(A3), isn't it can be calculated through the given condition, like below?
Pr(A1) + Pr(A2) + Pr(A3) = Pr(A1) + 1.5Pr(A1) + 2Pr(A1) = 4.5Pr(A1) = 1
Then Pr(A1) = 0.(2), Pr(A3) = 2Pr(A1) = 0.(4)
So why it's related to the number 0.23?
For number 2, I did it but I'm not sure, as below:
Pr(B1) = Pr(B1|A1)*Pr(A1) + Pr(B1|A2)*Pr(A2) + Pr(B1|A3)*Pr(A3)
With according to the table Pr(B1|A1) is 0.2, Pr(B1|A2) is 0.40... (If I'm not mistaken), and Pr(A1), Pr(A2), Pr(A3) can be found like in question 1.
For number 3, with Pr(B1), Pr(B2), Pr(B3) found in the same way as question 2, we calculate Pr(B1) + Pr(B2) + Pr(B3) - Pr(B1)Pr(B2) - Pr(B2)Pr(B3) - Pr(B3)Pr(B1) + Pr(B1)Pr(B2)Pr(B3). Is it true? I'm not really confident.