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You are given n=10,000 light bulbs. Each has a reliability of p=99.99% Suppose you select a batch of r=190 light bulbs to light your warehouse. Can you use the equation to predict the probability of having 189 good light bulbs out of this batch of 190?

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I am confused here as to what is considered the number of trials and number of successes (given the assignment of variables)...

Any point in the right direction is appreciated.

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up vote 1 down vote accepted

We need to make an interpretation: is exactly $189$ intended, or is it at least $189$?

We take the first interpretation, though there is no strong argument for it. The probability that a bulb is bad is $0.0001=p$. So the probability of exactly one bad out of $190$ is given by $$\binom{190}{1}p(1-p)^{189}.$$

Another way: This is a standard example of a setting where the Poisson approximation to the binomial is excellent. (If you have not covered the Poisson, then use the first solution.)

Here we have $n=190$. Let $\lambda=np$. The probability of one bad is approximately. $$e^{-\lambda}\frac{\lambda^1}{1!}.$$ In this case, the approximation is essentially dead on.

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thank you, is the 10,000 negligible or why is it negligible? – Greg McNulty Sep 14 '13 at 6:39
We could make two interpretations (i) probability an individual bulb is bad is $0.0001$ or (ii) there is exatly one bad among the $10000$. Interpretation (i) is more reasonable. But even if we take interpretation (ii), $190$ bulbs is a small fraction of the $10000$, so sampling with replacement (binomial) or without (hypergeometric) give essentially the same answer – André Nicolas Sep 14 '13 at 6:51

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