# 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?

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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?

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I guess it wouldn't be 100% chance of failure cause nothing ever is, but I would expect it to fail. It makes sense that it wouldnt scale past 100% if you had 11 batteries. –  Josh May 3 '11 at 19:18
@Josh, yes, that's an even better way to look at it. –  yunone May 3 '11 at 19:19
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.