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?

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
$$1-\left(\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}-5\left(\frac{1}{5}\right)^{14}\right).$$
