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In order for the circuit to work, current must be able to pass from left to right. Each component (A, B, C, D) has an independent probability of failure. Component A has a probability of failure of 0.03. Component B has a probability of failure of 0.04. Component C has a probability of failure of 0.05. Component D has a probability of failure of 0.06.

Given the circuit does not work, what is the probability that components C and D have failed? (Solution says $0.02804$)

I have attempted the question, and found the probability to be 0.7143 however, the solution lists it to be 0.02804, I'm not sure if I have made a mistake, here's how I solved it:

$$P(W_1^c) = P(A \cap B) = P(A)P(B) = 0.0012$$ $$P(W_2^c) = P(C \cap D) = P(C)P(D) = 0.0030$$ $$\text{I understand that I need to find} P(W_2^c | W^c)$$ $$P(W^c) = P(W_1^c \cup W_2^c)$$ $$P(W^c) = P(W_1^c) + P(W_2^c) - P(W_1^c)P(W_2^c)$$ $$P(W^c) \approx 0.0041964$$ $$P(W_2^c | W^c) = \frac{P(W_2^c)}{P(W_1^c) + P(W_2^c)}$$ $$P(W_2^c | W^c) = \frac{0.0030}{0.0042}=\frac{30}{42}\approx0.714285714 \neq 0.02804$$

Y K
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2 Answers2

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There are a few issues with your solution.

First, as I read it, you write that the event of failure $W^c$ is decomposed as either the entire top row failing ($W_1^c$) or the entire bottom row failing ($W_2^c$), which is not true. For instance, if the top row fails, but both $C$ and $D$ work, then the circuit is complete.

Instead, think of it like this: the circuit works if either $A^c$ and $B^c$ (corresponding to the top row functioning) or $C^c$ and $D^c$ (corresponding to the bottom row functioning). So, $P(W)=P((A^c\cap B^c)\cup(C^c\cap D^c))$ and hence $P(W^c)=P((A\cup B)\cap(C\cup D))$.

Secondly, your definition of conditional probability is a bit off. In general for events $E_1, E_2$ we have $P(E_1\mid E_2)=\frac{P(E_1\cap E_2)}{P(E_2)}$, so here it should be $P(W_2^c\mid W^c)=\frac{P(W_2^c\cap W^c)}{P(W^c)}$. Notice that $W_2^c\cap W^c=W_2^c\cap(A\cup B)$ (since if the bottom row have already failed, we just need one from the top to fail for the whole system to fail).

Using this, I get the result in your solution list.

muldyr
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Many conditional probability questions can be solved much nore simply, using the concept of reduced sample space.
This is a typical case of that type.

  • $P(W_2) = (1-0.05)(1-0.06) = 0.813$
  • $P(W_2^c) = 1 - 0.813 = 0.107$
  • $P(C \cap D) = 0.05*0.06 = 0.003$

That's all we need to compute

$P(C\cap D) | W^c) = \dfrac{P(C \cap D)}{P(W_2^c)}$

$$ = \frac{0.003}{0.107} = \frac{3}{107}, \approx 0.02804$$

That is the power of using the reduced sample space concept.
We didn't even bother about branch A-B,
because we know that it must be broken since the system has failed.

If you must, for completeness you can ascribe a value x for $W_1^c$ and multiply both the numerator and denominator by x

true blue anil
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