Originally Posted by

**awkward** I'm going to disagree with Mr. F here; X is the number of toys of a particular type, say toy #1, so he has computed the probability that toy #1 will not be found more than once in a group of 5, whereas the original problem asks that all the toys be distinct.

To that end, let m be the number of available toys, as before. The probability that the first toy taken is distinct, i.e. not a duplicate, is 1. The probability that the second toy is distinct is (m-1)/m because there are m-1 choices left. The probability that the third toy is distinct is (m-2)/m. ... Finally, the probability that the fifth toy is distinct is (m-4)/m.

So the probability that all toys are distinct is

$\displaystyle f(m) = 1 \cdot \frac{m-1}{m} \cdot \frac{m-2}{m} \cdot \frac{m-3}{m} \cdot \frac{m-4}{m}$.

We want the least m such that f(m) > 0.9. I get f(97) = 0.9006.