I've been all the morning searching for this and I ended with nothing clear... So I'm studying for my statistics exam an one excercise asks this:
If we have four computers A, B, C and D, whith each one of them having a chance to fail independant of the others( P(A) = 0'1; P(B) = P(C) = P(D)= 0,2 ). What is the probability of exactly one of them failing?
I've been looking here and there and ended with nothing. I found that in case of having exactly one of two uses the formula: P(A)+P(B)-2P(A)·P(B) but, i can't see how to escalate this to exactly one out of N (being N = 4 in this case).
Thank you everyone for your help :)
First calculate the probability of just A failing and B,C, and D not failing, which is $0.1 \cdot 0.8 \cdot 0.8 \cdot 0.8$
Now do the same for B failing and the other not failing, and then for C and D
Finally, add those all up