1
$\begingroup$

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.

enter image description here

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?

$\endgroup$
2
  • $\begingroup$ Sorry, I should have clarified that. D is the system is function, and D' is the system is not functioning. But I'm going to add a note in my original post and change it to S and S'. $\endgroup$
    – Doadle
    Sep 26, 2017 at 16:36
  • $\begingroup$ Yeah, I figured ... $\endgroup$
    – Bram28
    Sep 26, 2017 at 16:37

1 Answer 1

Reset to default
1
$\begingroup$

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:

$$P(S'|2) = P(3' \cap 4') = P(3') \cdot P(4') = (1-P(3))\cdot(1-P(4)) = $$

$$(1-0.9)\cdot(1-0.8)=0.1\cdot0.2=0.02$$

$\endgroup$
11
  • 1
    $\begingroup$ I'll add the $P(S'|2)$ to my answer ... $\endgroup$
    – Bram28
    Sep 26, 2017 at 16:47
  • 1
    $\begingroup$ @Doadle OK, added the $P(S'|2)$ to my Answer. Also, while your Bayesian formula certainly works, it may be a bit easier to work with: $P(2|S') = \frac{P(S'|2)\cdot P(2)}{P(S')}$, and to compute $P(S')=1-P(S)$ $\endgroup$
    – Bram28
    Sep 26, 2017 at 16:55
  • 1
    $\begingroup$ @Doadle You're welcome! :) $\endgroup$
    – Bram28
    Sep 26, 2017 at 16:55
  • 1
    $\begingroup$ @Doadle Correct! $\endgroup$
    – Bram28
    Sep 26, 2017 at 17:19
  • 1
    $\begingroup$ @Doadle No problem ... see how computing $P(S')$ was a little easier that way? $\endgroup$
    – Bram28
    Sep 26, 2017 at 17:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .