I am stuck with an exercise. It says we have a system consisting of two different kinds of components A and B with probabilites to fail of 10% and 20%, respectively. To build the system we use component A 4 times and component B 7 times. Now we are asked to find the probability that at least two components in the system are broken.
I see that we will have to calculate the probability of 0 components failing, as well as the probability of 1 component failing to get to what we want.
$P(0\; components \, fail)= 0.9^4 \cdot 0.8^7 = 0.14 $
However, I do not see how to deal with the fact that there are two different probabilites involved in the case P(1 component failes). Would it be some kind of average like $\frac{1}{2} \left( \binom{4}{1} 0.9^3 \cdot 0.1 \cdot 0.8^7 + \binom{7}{1} 0.9^4 \cdot 0.8^6 \cdot 0.2 \right) $ ?
Thank you very much already for your answers!