Bernoulli Trial variables?

Question

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?

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.

Answer 1

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.