Parallel circuit: probability of failure a) Relays used in the construction of electric circuits function properly with probability .9. Assuming that the circuits operate independently, what is the probability that current will flow when the relays are activated?

b) If we know that current is flowing, what is the
probability that switches 1 and 4 are functioning properly?

I denote $F_1$ the event switch 1 fails and $W_1$ the event switch 1 works and use the subscript $_S$ for the whole system and $1\leq i\neq j\neq k\neq l\leq4$ . Part a) is easy: this is $1-P(F_S)$ and $$P(F_S)=P(F_S|F_iF_jW_kW_l)P(F_iF_jW_kW_l)+P(F_S|F_iF_jF_kW_l)P(F_iF_jF_kW_l)+P(F_S|F_iF_jF_kF_l)P(F_iF_jF_kF_l)\\
=4.1^2.9^2+4*.1^3*.9+1^4=.9639$$
(which, the back pages of the book confirms, is the correct answer). Now, for part b) I think there are 7 combinations that allow the current to flow:
$$\{1:W_2W_4F_1F_3\}\\
\{2:W_1W_3F_2F_4\}\\
\{3:W_1W_2W_3F_4\}\\
\{4:W_1W_2W_4F_3\}\\
\{5:W_1W_3W_4F_2\}\\
\{6:W_2W_3W_4F_1\}\\
\{7:W_1W_2W_3W_4\}$$
Of these only combination 7,5 and 4 have $W_1W_4$ in them so that
$$P(W_1W_4??|F_S)=\frac{2*.1*.9^3+.9^4}{.9639}\approx.831$$
And when I do a classical approach using the Bayes formula I get to the same number. However the book says the correct answer is $.916$. 
I must add, I have done 40 exercises off this book already and it would be the first time one of the answers is incorrect. The book is very well edited. Therefore my guess is that I misunderstand the question. 
So, here is my question: what is my mistake?
 A: I got the same answer, using a probability tree.
A simulation supports it, as in this R example.
N <- 1e6
set.seed(17)
x <- matrix(runif(4*N) < 0.9, nrow=4)
flows <- (x[1,] & x[3,]) | (x[2,] & x[4,])
on.1.4 <- x[1,] & x[4,]

mean(flows) # Should approximate 0.9639
sum(flows & on.1.4) / sum(flows)

The output from these million iterations (in which 963,956 evidenced a flow) is
[1] 0.963956
[1] 0.8320089

The first number is comfortably close to $0.9639$ while the other clearly shows $0.832$ is plausible while $0.916$ is not.  The standard error in the second calculation will be approximately
$$\sqrt{0.832(1-0.832)/963956} \approx 0.00019,$$
showing the simulated estimate of $0.8320089$ is within $0.4$ standard errors of $0.9^3(1 + 2(0.1)) \approx 0.83193$.

Because of the special nature of the probabilities involved, we can also exhaustively enumerate the possibilities, as in this code:
p <- c(rep(1,9), 0)
X <- as.matrix(expand.grid(G1=p, G2=p, G3=p, G4=p))
flows <- (X[,1] & X[,3]) | (X[,2] & X[,4])
on <- X[,1] & X[,4]

sum(flows & on)
sum(flows)

The output shows the completely accurate answer is $8019/9639 = 0.8319328\cdots$.

Of course, the computer can easily enumerate all $16$ possibilities in the probability tree.  But you and I already did that manually, with the same results.
A: Looks like it's just that the second part of the problem is a poorly worded question. Instead lets rewrite it as two questions: "If we know that current is flowing, what is the probability that switch 1 is functioning properly? If we know that current is flowing, what is the probability that switch 4 is functioning properly?"
You can then solve using your same approach and using the 7 combinations you've outlined. For switch 1, the first of our two new questions corresponds to the combinations 2, 3, 4, 5, and 7. For switch 4: 1,4,5,6,7.
Both of these sets of combinations have the same probability:
$$Pr(W_1???|F_S) = Pr(W_4???|F_S) = \frac{.9^2*.1^2 + 3*.1*.9^3 + .9^4}{.9639} \approx 0.916 $$
I've got the same book on my shelf. Maybe they'll fix it in the eighth edition. 
