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 of one another and $P(\text{component works})=0.9$, calculate $P(\text{system works})$.
My try :
Call $1,2$ subsystem $A$, and $3,4$ subsystem $B$
\begin{align*}
P[\text{system fails}] &= P[A\text{ fails}] + P[B \text{ fails}] – P[\text{both fail}] \\
&= (0.1)^2 + (1 – (0.9)^2) – (0.1)^2 (1-(0.9)^2) \\
&= 0.1981
\end{align*}
Thus
$$P[\text{system works}] = 1 – 0.1981 = 0.8019$$
Best Answer
I am wondering why this doesn't work.
Let $A$, $B$ be the events that these subsystems work, $A_i,B_i$ for $i=1,2$ be the event that component $i$ works, and I will use a bar to show that something doesn't work. Then, assuming the system will work if at least one subsystem works, \begin{align*} P(\text{System works})&=P(A\cup B)\\ &=P(A)+P(B)-P(A)P(B)\\ &=P(A_1\cup A_2)+P(B)-P(A)P(B)\\ &=P(A_1)+P(A_2)-P(A_1)P(A_2)+P(B_1)P(B_2)-P(A)P(B)\\ &=.9+.9-(.9)^2+(.9)^2-P(A)P(B)\\ &=0.99+ 0.81-(.99)(0.81)\\ &=0.9981 \end{align*}