Lifetime of a light bulb Light bulbs average 800 hours of life (here the life time of a bulb follows an $\exp(1/800)$ distribution). Buy 1000 bulbs. Find the probability that the fourth bulb has gone out before 3 hours have passed.  
 A: The probability that a specific bulb dies in time $\le 3$ is $1-e^{-3/800}$. Call this $p$. 
Then the probability that the fourth shortest-lived bulb has life $\le 3$ is $1$ minus the probability that $3$ or fewer bulbs are dead by that time. 
This can be calculated by the standard formula for the binomial distribution. 
We can alternately use the Poisson approximation to the binomial, or by direct use of the Poisson. The mean number of "deaths" in $3$ hours is $1000\cdot\frac{3}{800}$. Let $\mu$ be this number. Then the probability of $k$ deaths in $3$ hours is approximately $e^{-\mu}\frac{\mu^k}{k!}$. Calculate the sum of the probabilities, $0$ to $3$, and subtract the sum from $1$.
Remark: Lifetimes of the bulbs, at least if they are installed in the same building, are not independent. Think power surge!
Also, the exponential is not a really good model for bulb lifetime. There is a peak of early deaths, because of manufacturing defects. And the exponential distribution is appropriate for things that die but do not age, like radioisotopes. Lightbulbs age. 
