# Probability of failure of an appliance. Total probability?

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.

• 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 ? – justt Jun 26 '13 at 11:52

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 $$P_F(A' \cup C') = \frac{P((A' \cup C') \cap F)}{P(F)} = \frac{P(A' \cup C')}{P(F)}$$

• Is is true that $P(A'\cap B' \cap C') = P(A')P(B')P(C')$? If so, why? Are they independent? – Spine Feast Jun 26 '13 at 12:22
• 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. – justt Jun 26 '13 at 12:24
• Ok, thanks for your answer. I keep overthinking these problems! – Spine Feast Jun 26 '13 at 12:35

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 $1$