Let's say I have a set of 500 fruits and 5 of them are kiwis (1%).
If I randomly take a sample of 50 fruits from the initial 500 set, what's the probability of having at least 1 kiwi ?
I suspect I should use the Stirling numbers of the second kind and also, because of the "at least", I imagine it'd be easier to calculate the probability of not having kiwis and then calculate the probability of having at least 1 kiwi as: $$ Answer = 1 - P(no kiwi)$$
I also know that there are $\left( \frac{500}{50} \right)$ possible samples of size 50 but I cannot put all the pieces together :(
Can you help me ? Many thanks in advance for your time !
P.S. I expect this probability to be low but while playing with numbers I end up with something unreasonably small (e.g. ~$10^{-50}$). It is hard also to calculate $\frac{500!}{50!}$... How do you do it ? I cannot multiply by hand $ 500*499*...*51 $, can I ? I guess the correct answer doesn't involve numbers with >1000 digits
Thanks to the comments I can answer my own question. It is actually much easier than I expected (I just messed up the calculations). Forget about the Stirling numbers, as I said it is much easier to calculate the probability of having at least one kiwi, $P(kiwi)$, as: $$ P(kiwi) = 1 - P(no\_kiwi) $$ where $P(no\_kiwi)$ is the ratio between the number of ways to pick 50 fruits without kiwis, $\binom{495}{50}$, and the total number of ways of randomly pick 50 fruits, $\binom{500}{50}$.
Explicitly: $$ \require{cancel} \begin{aligned} P(no\_kiwi) &= \frac{\binom{495}{50}}{\binom{500}{50}} \\ &= \frac{495!}{50!(495-50)!}*\frac{50!(500-50)!}{500!} \\ &= \frac{495!}{\bcancel{50!}*445!}*\frac{\bcancel{50!}*450!}{500!} \\ &= \frac{\bcancel{495!}}{\bcancel{445!}}*\frac{450*449*448*447*446*\bcancel{445!}}{500*499*498*497*496*\bcancel{495!}} \\ &= \frac{18045929260800}{30629362512000} \\ &\simeq 0.59 \end{aligned} $$ The probability we are looking for is therefore: $$ P(kiwi) = 1 - 0.59 \simeq 0.41 = 41 \% $$