The following circuit operates if and only if there is a path of functional devices from left to right. The probability that each device functions is as shown. Assume that the probability that a device is functional does not depend on whether or not other devices are functional.
What is the probability that device A
a) does not work if the system does not work?
b) does not work if the system operates?
So I have tried working on finding the $\mathbb{P}(A|S)$, wherein $\mathbb{P}(S)$ is the probability that the system will not work and $\mathbb{P}(A)$ is the probability that device A will not work. However, I just can't seem how to find $\mathbb{P}(A\cap S)$.
From what I computed, $\mathbb{P}(S)=0.070625$.
How can I find $\mathbb{P}(A\cap S)$?
Edit: there was a first problem about this one and by solving that I already knew what to do with P[S]. Btw, thank you for the edits people hehehe
I got a result close to yours as of $P(S)$.
I calculated first
$$P(work)=0.9\cdot0.8\cdot0.7[1-0.95^3]+0.95^3[1-0.9\cdot0.8\cdot0.7]+0.9\cdot0.8\cdot0.7\cdot0.95^3=0.929258$$
Thus your $P(S)=0.070742$. Perhaps your calculation are ok but there is just a different approx.
Now to calcualte $P(\overline{A} \cap S)=0.1\cdot[1-0.95^3]=0.0142625$
(if A doesn't work, in order to have the system out of work it is necessary that the lower branch of the system doesn't work too)
Thus $P(\overline{A}|S)\approx 20.16\%$
Similar arguments for the other request
EDIT: the easiest way to calculate $P(S)=[1-0.9\cdot0.8\cdot0.7][1-0.95^3]=0.070742$...I did a lot of useless calculations... :(