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Binomial (89,k) does indeed correspond to $\binom {89} k.$ We start with an easier problem: what is the probability of not getting $k$ specific toys in $225$ boxes? Since there are $90$ possible toys and each box is independent, this is just $(1-\frac{k}{90})^{225},$ the probability of not getting one of our specific k toys $225$ times in a row. Now, we ...


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