Probability

Hypergeometric Distribution


圖片參考：https://s.yimg.com/lo/api/res/1.2/U.hWrTAE333r8f9neLO4.w--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/http://upload.wikimedia.org/wikipedia/commons/thumb/e/ed/Raccoon_%28Procyon_lotor%29_2.jpg/220px-Raccoon_%28Procyon_lotor%29_2.jpg


61 racoons (浣熊)
★ 8 are tagger
★ 61 - 8 = 53 are untagged

For 20 captured raccoons, the probability that 4 of them are tagged is
(₈C₄)(₅₃C₁₆)/(₆₁C₂₀) = 0.166615228


2015-02-12 01:36:41 補充：
Awesome program.

Thanks Andrew for providing an empirical probability approach.

Cheers!

╭∧---∧╮
│ .✪‿✪ │
╰/) ⋈ (\\╯

Awesome.

I spent the morning thinking about it, come up with a program that computes exactly the same value using recurrence relations.

http://pastebin.com/R4LNhcPD

2015-02-12 16:56:45 補充：
The approach is also analytic - in the sense that it computes the exact correct probability value.

2015-02-12 16:57:09 補充：
We could be empirical as well - trying to do some Monte Carlos simulations (e.g. play the capture raccoons game 1 million time and measure the empirical probability). We will not get an exact answer in this case, but it is quick and dirty to get to an approximate solution in no time :)