Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that subsystem works iff either 1 or 2 works; since 3 and 4 are connected in series, that subsystem works iff both 3 and 4 work. If components work independently of one another and $P(\text{component works})=.9$, calculate $P(\text{system works})$.
Remark: The exercise is copied from the reference book Devore2017B.
1
Let $A_i$ denote the event that the component $i$ works.
Then, the system works if and only if subsystem 1 works or subsystem 2 works. While subsystem 1 works iff $A_1$ or $A_2$ occurs. The subsystem 2 works iff both the events $A_3$ and $A_4$ occur. Therefore,
\[
\begin{split}
P(\text{system works})&=P\bigl[(A_1\cup A_2)\cup(A_3\cap A_4)\bigr]\\
&=P(A_1\cup A_2)+P(A_3\cap A_4)-P\bigl[(A_1\cup A_2)\cap(A_3\cap A_4)\bigr]\\
&=P(A_1)+P(A_2)-P(A_1\cap A_2)+P(A_3)\cdot P(A_4)-P\bigl[(A_1\cap A_3\cap A_4)\cup(A_2\cap A_3\cap A_4)\bigr]\\
&=P(A_1)+P(A_2)-P(A_1)\cdot P(A_2)+P(A_3)\cdot P(A_4)-\Bigl[P(A_1\cap A_3\cap A_4)+P(A_2\cap A_3\cap A_4)-P(A_1\cap A_2\cap A_3\cap A_4)\Bigr]\\
&=0.9+0.9-0.9\times 0.9+0.9\times 0.9-\bigl[0.9^3+0.9^3-0.9^4\bigr]\\
&=1.8-[2\times 0.9^3-0.9^4]\\
&=0.9981
\end{split}
\]