Probability of system of devices. A system consists of four components $\{1,2,3,4\}$. The reliability of each device respectively is $\{.9, .95, .95, .99\}$. There are two routes ${1,2,4}$ and ${3,4}$ to complete the system. 
(1) What is the probability that the system operates at the end of this time period?
(2) What is the probability that component no. 3 is down and the system still   operates?
(3) What is the probability that at least one of the four components is down?
(4) Given that the system is down, what is the probability that it will resume operating 
if we replace component no. 1 by an identical (nondefective) component?
My Attempts:
(1)  ???
(2) $\Pr = \Pr(1)\Pr(2)\Pr(4) = .84645$
(3) $(1 -\Pr(1)\Pr(2)\Pr(3)\Pr(4)) = .1959$
(4) ???
 A: We are expected to assume that failure of components $1, 2, 3, 4$ are independent events.  
Problem $1$: We need the probability that $1$, $2$, and $4$ work or $3$ and $4$ work.  
For the system to work, we need $4$ to work. This has probability $0.99$.
In addition, we need $1$ and $2$ to work, or $3$ to work, or both. Recall the formula $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$. 
Let $A$ be the event $1$ and $2$ both work. This has probability $(0.9)(0.95)$. Let $B$ be the event $3$ works. This has probability $0.95$. So 
$$\Pr(A\cup B)=(0.9)(0.95)+ 0.95-(0.9)(0.95)(0.95).$$
Multiply by the probability that $4$ works to get the answer to our problem. 
Problem $2$:  We need $3$ to be down and $1$, $2$, $4$ to be OK. This is the product of four terms. The answer is $(0.9)(0.95)(0.05)(0.99)$. In your calculation, you left out the term $0.05$.
Problem $3$: The method you used is correct.
Problem $4$: Let $D$ be the event the system is down, and let $A$ be the event that the system is down because of $A$ only. We want $\Pr(A|D)$. This as usual is $\frac{\Pr(A\cap D)}{\Pr(D)}$.
You know already $\Pr(D)$, since that's a close relative of your answer to problem $1$.
So now we want the probability the system is down and replacing $1$ will fix it. That is the case only if all four of the following conditions hold: $4$ is good, $3$ is bad, $2$ is good, and $1$ is bad.  Compute the appropriate product, and divide by $\Pr(D)$. 
