# probability of a system made by sub-components

Let say we have a system made by 4 independent components ($A_j, j=1 \dots 4$) with the same probability of failing of $P(A_j) = 0.3$, now the probabilitay of a working components is $P(\bar{A_j}) = 1 - 0.3 = 0.7$.

Questions:

1) probability of the system to work $P(S)$, which means all components should work:

$$P(\bar{A_1} \land \bar{A_2} \land \dots \land \bar{A_5}) = (1-0.3)^4 = 0.24$$

2) Now suppose that the system has a failure, now what's the probability of that only the component $A_j$ has failed? i could write it also as $$P(A_j|S) = \frac{P(S|A_j)}{\sum_{i=1}^{4} P(S|A_j)P(A_j)}$$ correct?

but how can i get $P(S|A_j)$? is that even correct to use bayesian theorem here?

thanks

You notation is a little bit confusing. If $S$ denotes the working system then the not working system should be denoted as $\overline S$ or $S^c$.

Using the Baysian theorem, the probability that $\texttt{only}$ the component $A_j$ has failed given the system has a failure is

$$P(A_j^o|S^c)=\frac{P(S^c\cap A_j^o)}{P(S^c)}=\frac{P( A_j^o)}{P(S^c)}$$

where $A_j^o$ denotes the event that the component $A_j$ fails $\texttt{only}$.

The probability that component $A_j$ fails only is $0.3\cdot 0.7^3$

And the probability that the system fails is $P(S^c)=1-P(S)$

Remark

The equation $P(S^c\cap A_j^o)=P( A_j^o)$ holds because $A_j^o$ is a subset of $S^c$:

$A_j^o\subset S^c$