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On a small island, 2 rabbits are cuaght.Each one is tagged

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On a small island, 2 rabbits are cuaght.Each one is tagged [#permalink] New post 11 Aug 2003, 08:44
On a small island, 2 rabbits are cuaght.Each one is tagged so that it can be recognized again and is then released. Next day 5 more(untagged) rabbits are cuaght. Each one is tagged and realeased. On 3rd day 4 rabbits are cuaght, two of them are found to be already tagged.

Assuming the rabbit population in the island is constant and that tagged or untagged rabbit is equally likely to be caught, Find

(1) the smallest number of rabbits there can be in the island

(2) the probability of 2nd day's catchif there are exactly 12 rabbits in the island.
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Answer to Part (1) [#permalink] New post 11 Aug 2003, 08:59
(1) The smallest Number is 9. Seven are tagged and we just found two more that are untagged.

(2) 7/22 (got this from 10/12 * 9/11 * 8/10 * 7/9 * 6/8)
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Dmitry [#permalink] New post 14 Aug 2003, 06:01
Dmitry wrote:
stolyar wrote:
agree on both


agree on both

Great work guys..Dmitry, I would like to see the "front" of that poster...
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Re: Dmitry [#permalink] New post 17 Aug 2003, 07:10
mciatto wrote:
Great work guys..Dmitry, I would like to see the "front" of that poster...


mciatto, i assure you there is nothing interesting in front of that poster. :rotate
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 [#permalink] New post 20 Aug 2003, 18:41
The trick is to realize that on 2nd day , there are 10 untagged and 2 tagged rabbits..

Well Done Everyone !!
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Re: Rabbits - Part 1 [#permalink] New post 20 Aug 2003, 19:52
Brainless wrote:
On a small island, 2 rabbits are cuaght.Each one is tagged so that it can be recognized again and is then released. Next day 5 more(untagged) rabbits are cuaght. Each one is tagged and realeased. On 3rd day 4 rabbits are cuaght, two of them are found to be already tagged.

Assuming the rabbit population in the island is constant and that tagged or untagged rabbit is equally likely to be caught, Find

(1) the smallest number of rabbits there can be in the island

(2) the probability of 2nd day's catchif there are exactly 12 rabbits in the island.


I agree with the answer to question 2.

For question 1, 1 think 14 is the answer.

On third day, we know there are 7 tagged rabits (2 +5) out there. Of the 4 rabits that are caught 2 are tagged and 2 are not. If we assume the rabit population in the island is constant and that tagged and untagged rabit is equally likely to be caught, then the smallest number of rabits in the island must be 14 (2 * 7).

If we assume the smallest number of rabits in the island is 9 (7 + 2, that we have caught so far) then the probability of catching 2 untagged rabits out of 9 is not "equally likely" (skewed). Do I read too much out of it :?: :roll:
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 [#permalink] New post 20 Aug 2003, 23:35
i do not follow.could you please explain how you got the probabilities?

:(
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 [#permalink] New post 20 Aug 2003, 23:58
I see flaws in the statement

1. If we assume that tagged or untagged rabbit is equally likely to be caught the number of rabbits on island HAS to be 14 as during 2 days we have already tagged 7 rabbits. So, there are no smallest or greatest numbers - only 14 rabbits.
2. What kind of probability are we looking for? What catch is meant?
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Re: Rabbits - Part 1 [#permalink] New post 21 Aug 2003, 05:51
bnagasub wrote:
Brainless wrote:
On a small island, 2 rabbits are cuaght.Each one is tagged so that it can be recognized again and is then released. Next day 5 more(untagged) rabbits are cuaght. Each one is tagged and realeased. On 3rd day 4 rabbits are cuaght, two of them are found to be already tagged.

Assuming the rabbit population in the island is constant and that tagged or untagged rabbit is equally likely to be caught, Find

(1) the smallest number of rabbits there can be in the island

(2) the probability of 2nd day's catchif there are exactly 12 rabbits in the island.


I agree with the answer to question 2.

For question 1, 1 think 14 is the answer.

On third day, we know there are 7 tagged rabits (2 +5) out there. Of the 4 rabits that are caught 2 are tagged and 2 are not. If we assume the rabit population in the island is constant and that tagged and untagged rabit is equally likely to be caught, then the smallest number of rabits in the island must be 14 (2 * 7).

If we assume the smallest number of rabits in the island is 9 (7 + 2, that we have caught so far) then the probability of catching 2 untagged rabits out of 9 is not "equally likely" (skewed). Do I read too much out of it :?: :roll:


Equally Likely outcomes : we all agree that unless the events are equally likely , we can not calculate the the theoritical probability .. That means, in our case, each of TAGGED or UNTAGGED rabbit has an equal chance of to be caught .. ( here the event is CATCHING A rabbit). And as you said, probabilty of catching a 2 UNTAGGED rabbits = 2 * 1/9 , as u may catch any of them..

Well , I too felt , that the wording of the question was NOT upto ETS Standard...
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 [#permalink] New post 21 Aug 2003, 11:08
bono wrote:
I see flaws in the statement

1. If we assume that tagged or untagged rabbit is equally likely to be caught the number of rabbits on island HAS to be 14 as during 2 days we have already tagged 7 rabbits. So, there are no smallest or greatest numbers - only 14 rabbits.
2. What kind of probability are we looking for? What catch is meant?


When I toss a coin the outcomes are assumed to be equally likely. Because
p(H) = p(T) = 1/2 [ or assumed to be very close to 1/2]. If we do an empirical test of 100 tosses and found 48H and 52T, we still would consider the coin fair.
If the outcomes were 80H and 20T then the coin would be considered skewed and the outcomes of tossing the skewed coin are not equally likely. Probability of getting a head is no longer 1/2, it is approximately 4/5.

In the rabbit example, we know we have 7 tagged rabbits in the island. If we assume there are only 9 rabbits in the island then
p(untagged) = 2/9
p(tagged) = 7/9
and the result of any empirical test, of catching the rabbits, would be skewed toward the tagged rabbits. [not equally likely]

If we have 14 rabbits then p[U] = p[T] = 1/2.

If we assume we have only 13 rabbits, we know for a fact the probability of any catch is a little bit skewed toward tagged rabbits. We may consider the outcomes equally likely. But we know they are not.

If we have 15 rabbits then p[U] and p[T] are approximately equal to 1/2. We still may consider the catch as equally likely and we know the outcomes are skewed a little toward untagged.

Hence, 14 is the smallest number of rabbits there can be in the island, to be sure that a catch is equally likely.
  [#permalink] 21 Aug 2003, 11:08
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