I'm absolutely blanking this.

There are three devices to make the circuit. Device 2 and 3 are in series, but device 1 is in parallel to device 2 and 3. I have a probability value for 1, 2, and 3. Device 1 is independent to device 2 and 3. However, if device 3 fails, then device 2 has a conditional probability to fail.

My task is to find the probability the entire circuit will fail. The values are below.

$P(E_1) =0.2, P(E_2)=0.2, P(E_3)=0.3, P(E_2|E_3)=0.5$

I recall if they were all independent then I'd simply multiply the three probabilities together, however as device 2 and 3 are not independent to each other I recall something has to be done. But I'm stump at this part. Any documents regarding conditional probability focuses on $P(E_2|E_3)=\frac{P(E_2 \cap E_3)}{P(E_3)}$, but doesn't expand on what should be done beyond this.

Anyone have any resources or help regarding this?


Also, how would you draw the Venn diagram? I'm trying to visualize it in my head to make it easier to understand, but the diagram is not coming to me. My current diagram has 1 and 3 intersecting as a larger circle with 2 being the "smaller" circle within 3, however, I feel this is incorrect.


1 Answer 1


I'm not sure whether the entire circuit fails if any component fails, or only if all the components fail.

However, let's consider how to deal with the joint probabilities for $E_2$ and $E_3$. The Venn diagram is indeed a simple way to proceed. Let us consider that

$$ P(E_2 \mid E_3) = \frac{P(E_2, E_3)}{P(E_3)} $$

That means that if we draw a Venn diagram for just those two events, the region for $E_3$ (orange, below) covers an area of $0.3$, and the region for $E_2$ (blue) covers an area of $0.2$, and the region of overlap is half the area for $E_3$ (because $P(E_2 \mid E_3) = 0.5$), or $0.5 \times 0.3 = 0.15$. That leaves $0.2-0.15 = 0.05$ for the area of the region of $E_2$ that lies outside $E_3$.

enter image description here

In summary, we thus have

  • $P(E_2, E_3) = 0.15$
  • $P(E_2, \neg E_3) = 0.05$
  • $P(\neg E_2, E_3) = 0.15$

which leaves

  • $P(\neg E_2, \neg E_3) = 1-0.15-0.05-0.15 = 0.65$

Thus, if any component failure is sufficient to have the entire system fail, then the desired probability is one minus the probability of all components surviving, or $1-[1-P(E_1)]P(\neg E_2, \neg E_3)$. If, on the other hand, all the components must fail for the entire system to fail, then the desired probability is $P(E_1)P(E_2, E_3)$.

There are other interpretations of system failure, which I trust you can take from here.

  • $\begingroup$ Wow thanks! I completely forgot that because $E_1$ is independent to event 2 and 3 that as far as a Venn diagram is concerned the circle for event 1 won't even be touching the other two circles. Thanks! $\endgroup$
    – user323111
    Mar 15, 2016 at 18:16
  • $\begingroup$ No, it's not that it doesn't touch. It's not in the diagram. We are only concerned about the interaction between $E_2$ and $E_3$. $\endgroup$
    – Brian Tung
    Mar 15, 2016 at 18:22
  • $\begingroup$ Ahh ok then. Thanks. So if the following is within the set of failure, $E_fail = {E_1 E_2, E_1 E_3, E_1 E_2 E_3}$, then to find the total probability of system failure I need to do $P(E_1)P(E_2)+P(E_1)P(E_3)+P(E_1)P(E_2,E_3)$ right? Also, the comma between them are multiplications correct? $\endgroup$
    – user323111
    Mar 15, 2016 at 18:29
  • 1
    $\begingroup$ No, because you've triple added the cases where all three components fail. The actual expression should be $P(E_1)[P(E_2)+P(E_3)-P(E_2,E_3)]$, by inclusion-exclusion. $\endgroup$
    – Brian Tung
    Mar 15, 2016 at 18:47

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