標題:
Probability
發問:
A park has 61 raccoons of which 8 were previously captured and tagged. Suppose that 20 raccoons are captured. What is the probability that 4 of these are found to be tagged?
最佳解答:
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 (?C?)(??C??)/(??C??) = 0.166615228 2015-02-12 01:36:41 補充： Awesome program. Thanks Andrew for providing an empirical probability approach. Cheers! ╭∧---∧╮ │ .??? │ ╰/) ? (\╯
其他解答:
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 :)
Probability
發問:
A park has 61 raccoons of which 8 were previously captured and tagged. Suppose that 20 raccoons are captured. What is the probability that 4 of these are found to be tagged?
最佳解答:
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 (?C?)(??C??)/(??C??) = 0.166615228 2015-02-12 01:36:41 補充： Awesome program. Thanks Andrew for providing an empirical probability approach. Cheers! ╭∧---∧╮ │ .??? │ ╰/) ? (\╯
其他解答:
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 :)
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