Probability of a machine with two switches failing A certain type of switch has a probability p of working when it is turned on. Machine A, which has two of these switches, needs both such switches to be switched on to work, and machine B (also with two of the same switch) only needs at least one of them switched on to work.
How can I calculate the probability that each machine will fail to work?
 A: Assuming that the switches fail or do not fail independently the probability of both failing (for machine B) is the product of the probabilities: $p^2$.  The probability that either switch fails (for machine A) is the complement of the probability that neither fails, or $1 - (1 - p)^2 = p(2 - p)$.
So the probability of A failing is $p(2 - p)$ (either switch fails) and the probability of B failing is $p^2$ (both switches fail).  You can check that when $0 < p < 1$ the probability that A fails is greater than the probability that B fails, which is intuitively correct since the condition that makes A fail is "easier" than the condition that makes B fail.
A: Let $ q=1-p $ be the probability of a switch failing and let $ P(n) $ be the probability that, for a given machine, n switches succeed. For machine A to fail, you need to find $ P_a( \leq 1)=P_a(1)+P_a(0) $ because as long as one switches fails, the whole thing does. For machine B to fail, you need $ P_b(0) $ as both switches must fail. Assuming that the success of one switch does not affect the probability for the other, you can invoke the product rule; that is, the probability of an event for one switch and an even for another coinciding is the product of their probabilities. From there, hopefully you can figure out the rest.
A: Prob of both switches on A working: $p^2$
Prob of one of switches on B working = prob NOT both of them fail: $1-(1-p)^2$
so $p^2(1-(1-p)^2)$ is the probability both is working
