Statistical Probability We make sweets of $5$ different flavors.  The sweets are made in batches of $1.8$ million sweets per batch and then combined/mixed together into a total random batch of $9$ million sweets.
We then pack into tubes of $14$ sweets and we would like to know the probability of getting all $5$ flavors in every pack.  If the sweets are fed to the packing machine completely randomly from the mixed batch, how do we calculate this?
 A: The details depend on whether you want a theoretically exact answer, or an excellent approximation. Since this is packaged as an applied problem, we work with the approximation. So we will assume that each sweet has one of $5$ values, independently of the values of the previously chosen sweets. The number $9,000,000$ is so large that "with replacement" is essentially equivalent to "without replacement."
Let the flavours be called $1,2,3,4,5$. We find the probability that some flavour(s) is (are) missing. The method we use is Inclusion/Exclusion.
First we find the probability that flavour $1$ is missing. That is the probability all $14$ sweets are not of type $1$. Since the probability of being of type $1$ is $\frac{1}{5}$, the probability the sweets are all of type other than $1$ is $\left(\frac{4}{5}\right)^{14}$.
Now add together the probabilities for the $5$ flavours. We get a first estimate 
$5\left(\frac{4}{5}\right)^{14}$ for the probability that some flavour(s) is (are) missing. 
But for every pair $\{i,j\}$ of flavours, we have counted twice the probability that flavours $i$ and $j$ are both missing. The probability $i$ and $j$ are both missing is $\left(\frac{3}{5}\right)^{14}$. We must add up over the $\binom{5}{2}=10$ possible pairs, and subtract from our first estimate. So our next estimate is $5\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}$.
But we have subtracted too much, for we have subtracted twice the probability that three flavours are missing. The probability that the flavours $i,j,k$ are missing is $\left(\frac{2}{5}\right)^{14}$. Add over the $\binom{5}{3}=10$ triples. So our next estimate is
$5\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}$.
We have added a little too much, for we have added one too many times the probabilities that $4$ flavours are missing. Thus the probability at least one flavour is missing is $5\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}-5\left(\frac{1}{5}\right)^{14}$. Thus the probability no flavour is missing is
$$1-\left(\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}-5\left(\frac{1}{5}\right)^{14}\right).$$
