What approximately is the probability that she will lose 2 or more pieces of luggage According to an airline report, roughly 1 piece of luggage out of every 2000 that are checked is lost. Suppose that a frequent-flying businesswoman is checking 1200 bags over the course of the next year. What approximately is the probability that she will lose 2 or more pieces of luggage
 A: We need to make some assumptions. The first is that she is "typical," that the airlines she travels on have a probability $1/2000$ of losing any particular piece of luggage. We are also invited to make the highly dubious assumption of independence. It seems likely to me that if one checks two pieces of luggage, then given one of them is lost, the other has heightened probability of being lost. 
But let us hold our noses and go on. If $X$ is the number of pieces of luggage lost, then under the assumptions we have that $X$ has binomial distribution, $n=1200$, $p=1/2000$. We want the probability that $X\ge 2$. 
It is easier to find the probability of the complement, the probability that $X\le 1$. This is $\Pr(X=0)+\Pr(X=1)$. Recall that for the binomial, the probability that $X=k$ is $\binom{n}{k}p^k (1-p)^{n-k}$.
Another way: The random variable $X$ is binomially distributed, with $p$ small, $n$ large, and $np$ of moderate size. Under these circumstances, the binomial is well approximated by the Poisson with $\lambda=np$. In our case, $\lambda=1200/2000$.
The probability that $X=k$ is approximately $e^{-\lambda}\frac{\lambda^k}{k!}$. So
$\Pr(X\le 1)\approx e^{-\lambda}+e^{\lambda})\lambda)$.
