An electrical circuit has 5 components labelled A, B, C, D and E. The probabilities of components A, B, C, D and E working are 0.7, 0.7, 0.8, 0.8 and 0.8, respectively. What is the probability that:
-
(a) The system will be working.
-
(b) Component A will not be working given that the system is working.
-
(c) Component A will not be working given that the system is not working.
I solved but I'm not sure it is correct.
-
(a)
W is event that the system will be working
\begin{align}
& P(W) \\
={}& P(A \cap B \cup C \cap D \cap E)\\
={}& P(A \cap B) + P(C \cap D \cap E) – P(A \cap B \cap C \cap D \cap E)\\
={}& 0.7^2 + 0.8^3 – 0.7^2 0.8^3\\
={}& 0.75112
\end{align} -
(b)
\begin{align}
& P(\bar{A}|W) \\
={}& \frac{P(\bar{A} \cap W)}{P(W)}\\
={}& \frac{P(\bar{A}BCDE \cup \bar{A} \bar{B}CDE)}{P(W)}\\
={}& \frac{0.3\times0.7\times0.8^3 + 0.3^2\times0.8^3}{0.75112}\\
={}& \frac{1920}{9389}
\end{align} -
Is my answer for (a) and (b) correct?
-
How to solve (c)?
Best Answer
The figure gives $P(C)=0.8$ but the question says $P(C)=0.7$.
Your answers are correct according to the figure but there are some typos: in $(a)$ it should be $P(A\cap B\cup C\cap D\cap E)$ as taking union of probabilities does not make sense. In $(b)$, judging from your calculation and assuming $XY$ is shorthand for $X\cap Y$, you probably meant to have $P(\bar ABCDE\color{red}\cup\color{blue}{\bar A\bar B}CDE)$ in the numerator. Note that $\overline{AB}\ne\bar A\bar B$.