Conditional probability question? Suppose that in a network of 3 computers, at a given time the event that the kth computer is down has unconditional probability pk for k = 1, 2, 3. Moreover, there is probability p of power failure, in which case all the computers are down, but given that there is no power failure the computers are up or down independently of each other. Calculate the probability that at this time there is at least one computer up
 A: We assume that the quoted probabilities $p_1$, $p_2$, $p_3$ are probabilities when power is available. 
As is fairly often the case, it is easier to calculate first the probability that no computer is up. 
This can happen in two disjoint ways: (1) The power is down or (2) The power is not down, but all three computers fail.
The probability of (1) is $p$.
The probability of (2) is $(1-p)p_1p_2p_3$.
Thus the probability we cannot compute is $p+(1-p)p_1p_2p_3$, and therefore the probability that at least one computer is up is 
$$1-\left[   p+(1-p)p_1p_2p_3\right]             .$$
There are other ways to compute the probability. Almost the same as the previous argument is that we can compute precisely if the power is not down and at least one computer works. The probability at least one computer works is $1-p_1p_2p_3$, so the required probability is $(1-p)(1-p_1p_2p_3)$.
There are also messier ways to do the calculation, by computing separately the probability that exactly $1$ computer works, exactly $2$, exactly $3$, and adding. 
