30 Jul
2016
30 Jul
'16
1:43 p.m.
Since we have replacement, the probability that object i is *not* selected in k trial is (1-1/n)^k, so the mean number of distinct objects selected is n*(1-(1-1/n)^k). Victor On Fri, Jul 29, 2016 at 11:30 PM, David Wilson <davidwwilson@comcast.net> wrote:
Let set S contain n objects, each object having equal probability of being selected. At the end of k selections with replacement, what is the expected number of distinct objects that were selected?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun