Find the probability of a functioning circuit 
I keep getting that
$$P(\text{Circuit works}) = P(\geq\text{1 of subcircuit 1}) +P(\geq\text{1 of subcircuit 2}) +P(\geq\text{1 of subcircuit 3})  $$
$$= (0.9*0.1^2 + 0.9^2*0.1 + 0.9^3)*(0.95^2 + 0.95*0.05)*(0.99) = 0.770$$
But this is the wrong answer

I think you have to use the binomial distribution. 
$P(\text{A=at least one device functions in the first subcircuit})$
$=1-P(\text{no device function in the first subcircuit})$
So $P(A)=1-\binom 30 0.9^0 0.1^3$
