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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
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1 Answer 1

up vote 4 down vote accepted

Well if you need two circuits working for the product to operate, you would have a 0% chance of operation if you only included 1 circuit.

For the math's sake, there are a lot of different cases that count as 'product working', since you only need at least two circuits to work. It's easier to find the probability of product failure, since there are two options: one circuit working or zero circuits working. So we find the probability of failure and subtract that from one:

P(product works)=1-P(zero circuits working) - P(one circuit working)

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

This is 81% for two circuits and 97.2% for three. So you should need at least three circuits.

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I agree with your derivation. However, when I put n = 8 in the expression, I get 0.99999927 instead of the claimed 81%. –  Mick Mar 11 at 17:46
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