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A circuit has three major components, $A, B$ and $C$. A component $A$ operates independently of $B$ and $C$, the components $B$ and $C$ are interdependent. It is known that the component $A$ works properly $70\%$ of the time, component $B$$ 85\%$ of the time and $C$$99\%$ of the time. However, if component C fails, there is a $55\%$ chance that B will also fail. Assume that at least two parts must operate for the circuit to function. What is the probability that the circuit will work properly?

I think it is $0.7\times0.85\times0.99 + 0.7\times0.45\times0.01 + 0.7\times0.99\times0.15 + 0.85\times0.99\times0.3$ (all of them working $+ A$ and $B$ working $+ A$ and $C$ working $+ B$ and $C$ working )

Is it okay ? thank you

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  • $\begingroup$ No, there is a small mistake in your calculation. You must first evaluate the probability $X$ that $B$ fails when $C$ works. This is found by solving: $0.15 = 0.99 * X + 0.01 * 0.55$. Once you have found the value of $X$, you can proceed as in your computation. $\endgroup$ – M. Wind Apr 28 '15 at 21:14
  • $\begingroup$ Can you explain it in more detail.... I quite dont understand when C is working there is 85% that B is working as well $\endgroup$ – Plokijuh Apr 30 '15 at 18:51
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If the probabilities of the three components $A$, $B$, $C$ were independent of each other (in other words: uncorrelated), then it would indeed be correct to simply multiply the probabilities associated with $ABC$, $AB$, $AC$ and $BC$ working. Followed by adding up these four numbers, giving you the chance that least two components work.

However, as clearly stated in the text, the probabilities for components $B$ and $C$ are interdependent! This means that, before you can do a calculation as in the previous paragraph, you must first explore fully the interdependence of these probabilities.

Okay, let us do this. It is given that $C$ has a failure chance of $0.01$. Furthermore it is given that if $C$ fails, then the chance that $B$ fails is $0.55%$. This leads us to the conclusion that the probability that both $C$ and $B$ fail is equal to $0.01 * 0.55 = 0.0055$.

With a bit more effort we can derive that the probability that $C$ fails and $B$ works is equal to $0.0045$. The chance that $C$ works and $B$ fails is $0.1455$. The chance that both $C$ and $B$ work is $0.8455$. We now have the four values that determine the interdependence of $C$ and $B$.

Finally we can compute the overall probability that the system works. We get:

$$P = P(ABC) + P(AC) + P(AB) + P(BC)$$

$$P = 0.7 * (0.8455 + 0.1445 + 0.0045) + 0.3 * 0.8455 = 0.9498$$

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