Well, since she won't put any of them back into the box after eating them, it should be a Hypergeometric distribution.
You can replace it with a Binomial distribution if n/N is > 0.05 but I don't think it's the case here.
So you could set:
N = 30
m = 5
n = 16
And set k = 0, which would get you the chance that if she picks 16 chocolates from 30 (5 of which have rum), there will be 0 with rum among the 16 she picked.
It doesn't say anything about whether she will eat one with rum directly afterwards but neither does the task description IMO.
The important part here being that she doesn't put any of the chocolates back and the Hypergeometric distribution is the only discrete distribution we learned that works when you don't put back what you picked out of N.
Edit: Thinking about it again, CR may be right that it implies the 17th will be a rum chocolate nugget, in that case you will have to do it manually I guess. There may be a chance that you could use the geometric distribution as well because N is 30 or so but that seems a bit flaky/inaccurate because it assumes she puts them back after eating them.
Or you could calculate the chance that she eats 16 non-rum nuggets and then calculate the chance that when she picks 1 from the remaining 14 nuggets, it's a rum nugget, then combine the two. Which is how CR did it except you'd speed up the calculation for the first 16 a bit by using the Hypergeometric distribution I guess.
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