# Probability of failure

An electronic product contains $8$ circuits. The probability that any of the circuits is defective is $0.10$, and they are independent. The product operates only if at least two of the circuits are non-defective.

Question is: what is the probability that the product operates?

$P(\text{nonDefective}) = 1 - 0.10 = .9$

Given it takes two circuits working, I assume the probability is $.9$$^2$ or $.81$?

How many circuits should the product contain to be $90$% certain it will work? I believe this is just one

Would these answers be correct?

Thanks

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No, as shown by RussH's answer, they are not. It takes at least two circuits working for the product to work. $.9^2$ would give the probability that exactly two, specific circuits are working. But there are other cases: perhaps the forth and fifth circuits are the working ones, or perhaps the first four only, or perhaps all but the fifth... – David Mitra Jan 17 '13 at 23:18

$=1-.1^n-n*.1^{n-1}*.9$