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On a small island, 2 rabbits are caught. Each is tagged so [#permalink]
 04 Jan 2005, 19:39
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On a small island, 2 rabbits are caught. Each is tagged so that it can be recognised again and is then released. Next day 5 more (untagged) rabbits are caught. These too are tagged and released. On the third day, 4 rabbits are caught and two of them are found to be already tagged.
Assuming the rabbit population is constant and that tagged and untagged rabbits are equally likely to be caught, find:
(a) the probability of the second day's catch if there are exactly 12 rabbits on the island.
(b) the joint probability of obtaining both the second and the third catches if there are exactly 16 rabbits on the island.
I have the official answers, however, I am not sure how they were calculated.
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banerjeea_98, you are correct. Please explain. Thanks.
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SVP
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a) By second day you have 2 tagged and 10 untagged. So prob of getting 5 untagged is : 10C5 / 12C5 = 7/22
b) In this scenario we have 16 rabbits, so prob of second day catch is 14C5/ 16C5
By third day we will have 9 tagged and 7 untagged rabbits, so prob of third day catch is 9C2*7C2 / 16C4
Multiplying 2nd and 3rd day probs = (14C5/ 16C5) * (9C2*7C2 / 16C4)= 99/520
Hope this helps.
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Intern
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Of course - another combinations problem. I should have known. Thanks again.
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