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An appliance consists of three elements of type A, three of type B, four of type C. The appliance stops working if less than two elements of a kind are functional. In some interval of time $t$, the probability of failure of the components is:

$P(A)=0.1, P(B)=0.1, P(C)=0.2$

-What is the probability of the appliance failing in time $t$?

-Given that the appliance failed, what is the probability of the failure being a result of type A or C failing?

I'm not sure how to set up the calculations for this. I tried doing the first one with the law of total probability, (either the failure was caused by A, B or C - type), but these are not independent. Would appreciate hints on how to begin.

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    $\begingroup$ The appliance stops working if less than two elements of the same kind fail. It does not make sense ? If zero elements of the same kind fail, does it stop working ? $\endgroup$ – justt Jun 26 '13 at 11:52
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First question : Let $A'$ be the event "less than two elements of type $A$ are functional". Its probability can be computed easily using binomial law. Let $B'$ and $C'$ be defined in the same fashion.

The event $F=$"the appliance fails" is the event $A'\cup B' \cup C'$. You can compute its inverse event first, which is $\bar F = \bar A' \cap \bar B' \cup \bar C'$.

Second question :

The probability you are looking for is $$P_F(A' \cup C') = \frac{P((A' \cup C') \cap F)}{P(F)} = \frac{P(A' \cup C')}{P(F)}$$

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  • $\begingroup$ Is is true that $P(A'\cap B' \cap C') = P(A')P(B')P(C')$? If so, why? Are they independent? $\endgroup$ – Spine Feast Jun 26 '13 at 12:22
  • $\begingroup$ stricto sensu I can't say, it should have been precised in your problem. But I guess that the failure of$A$-elements does not affect the failure of $B$-elements, and so on... so $A',B'$ and $C'$ should be independent, yes. $\endgroup$ – justt Jun 26 '13 at 12:24
  • $\begingroup$ Ok, thanks for your answer. I keep overthinking these problems! $\endgroup$ – Spine Feast Jun 26 '13 at 12:35
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Q1: think about it in the following way: either A or B or C fail. What happens after any of them fails? We don't care, the appliance is already down. So I suggest doing it like this: probability of failure of every component is clearly binomial: $$ P(A)=\binom{3}{2}p_a^2(1-p_a)\\ P(B)=\binom{3}{2}p_b^2(1-p_b)\\ P(C)=\binom{4}{3}p_c^3(1-p_c) $$

This is because as soon as either 2 of A or B or 3 of C fail, the appliance is down. Assuming independence, use inclusion-exclusion theorem: $$ P(F)=P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P(C)+P(A)P(B)P(C) $$

Q2: use Bayes formula $P(F|A)P(A)=P(A|F)P(A)$ and the fact that $P(\cdot|F)$ is clearly $1$

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