The system diagrammed below will function if a path from A to B can be found through independently functioning components. Component i functions with probability Pi, i = 1, 2, 3, 4, where P1 = 0.7, P2 = 0.95, P3 = 0.9, P4 = 0.8
Find the probability that component 2 functions given that the system does not function.
So what I tried was:
So it's asking for $P(2|S')$, where S is the system is functioning, and S' is the system is not functioning.
I proceeded by doing $P(2|S')$ = $\left( \frac{P(2 \cap S')}{P(S')} \right)$
Then I figured since the components were independent, I could break the numerator up into $\left( \frac{P(2)P(S')}{P(S')} \right)$, which would let me cancel out P(S'), leaving me with P(2) = 0.95.
Is this the correct way of thinking about this, or would I need to use another property like Bayes?
Your assumption that $2$ and $S$ are independent is not correct: $2$ functioning or not functioning has a definite effect on there being such a path: with $2$ functioning there is more likely to be a path then with $2$ not functioning, and the same goes for the other components. So, while the events $1$, $2$, $3$, and $4$ are all independent of each other, neither of them are independent of $S$.
And yes, you'll need to use Bayes... which means, among other things, that you need to figure out $P(S'|2)$
Well, given that $2$ is working correctly, and thus the 'signal' getting to the 'middle', the probability of the system not functioning is the probability of both $3$ and $4$ not functioning (otherwise, the system would function), and since $3$ and $4$ are independent, we get:
$$P(S'|2) = P(3' \cap 4') = P(3') \cdot P(4') = (1-P(3))\cdot(1-P(4)) = $$
$$(1-0.9)\cdot(1-0.8)=0.1\cdot0.2=0.02$$