Simple probability question If there are 3 batteries each with 10% chance of failure, it seems that the probability of failure would be 3*10% or .30. However I calculate it to be 1-((1-.1)^3) = 0.271. Why is there a difference between these methods of calculating it? Different assumption?
 A: You are adding probabilities in the first case and multiplying probabilities in the second case. You are also assuming independence. Note that $$P(F) = 1-P(F')$$ $$ = 1-P(B_{1}' \cap B_{2}' \cap B_{3}')$$ where $F$ denotes failure and $B_1, B_2$ and $B_3$ denote the probability that a battery fails. 
A: Your latter calculation calculates the complementary probability of the event that none fail, which is the same probability that at least one battery fails, which is what you want. The first case you are merely adding probabilities, which makes no sense. Suppose you had $10$ batteries instead of $3$, each with a $10\%$ chance of failure. Would it make sense to say that there is a $10\cdot 10\%=100\%$ chance of failure in that case?
A: In the first case, you are assuming the probabilities are exclusive.  That is, if any one battery fails then the others will not.  This situation can be realized if they are connected in serial: in this case any one battery failing will disconnect the circuit and prevent the failure of the others.
In the second case, you are assuming the probabilities are independent.  This is the same as adding up the seven exclusive probabilities p1*p2*p3 + p1*p2*(1-p3) + p1*(1-p2)*p3 + (1-p1)*p2*p3 + p1*(1-p2)*(1-p3) + (1-p1)p2(1-p3) + (1-p1)*(1-p2)*p3, corresponding to the single case when they all fail, the the three cases when two of them do, and the three cases when exactly one of them does, with p1 = p2 = p3 = 0.1.  This describes the situation when they are connected in parallel: if any one of them fails the others continue to supply current.
