Technical control checks a series of 100 devices. Device has type A defect with the probability of $0.01$ and type B defect with probability of $0.02$ these two defects are independent (type A wont increase or decrease a chances of appearing type B defect and vice versa). They conclude that device if defected if it has at least on of these two defects. Find the probability that there are two to five defected devices.
Lets say that $X$ is the number of defected devices, we can see that Poisson distribution will do just fine here since $np_A=1$ and $np_B=2$, but thats not the problem.
We can say that there can be two devices with defect A and that can be easily calculated, however, it might happen that there are two devices defected with type B defect but it can also happen that BUT, it can also happen that there are one with A and one with B, question is, should this case need to be counted twice, since for example, 55. device might have defect A and 88. can have defect B and thats one case, since it can happen other way around should that be calculated twice or not?