Question

There are N birds in a population. The first day we captured and tagged 60, di 60. The following day we capture 50, d2=50. Of

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Answer #1

after first day :

d1 birds tagged

N-d1 birds not tagged

Probability that out of d2 birds captured D are already tagged

= (select D tagged birds out d1 tagged birds)*(select (d2-D) non tagged birds out (N-d1) non tagged birds) / (select d2 birds out of N birds)

= P(D | d1,d2,N) = [ (d1CD) * (N-d1)C(d2-D) ] / (NCd2)

{ nCx = n! / (x! * (n-x)!) }

P(D | d1,d2,N) = [ (d1CD) * (N-d1)C(d2-D) ] / (NCd2)

(PLEASE UPVOTE)

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