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A fisherman caught 4000 fish. He marked at least 1/4 of the

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A fisherman caught 4000 fish. He marked at least 1/4 of the [#permalink] New post 19 Dec 2004, 10:52
A fisherman caught 4000 fish. He marked at least 1/4 of the fish with a single mark. He also marked 400 fishes with two or more marks. What is the probability that the fisherman will catch a fish with a single mark among all the marked fishes?

I do not have the answer choices. Please explain your answer.
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 [#permalink] New post 19 Dec 2004, 12:40
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A fisherman caught 4000 fish. He marked at least 1/4 of the fish with a single mark. He also marked 400 fishes with two or more marks. What is the probability that the fisherman will catch a fish with a single mark among all the marked fishes?


Did you actually mean "at least" in the question. If that is the case, i dont think u can solve the problem,

Assuming the question meant exactly 1/4.
then Fishes with one mark = 1000.
fishes with 2 or more = 400.

so reqd probability = 1000/(1000 + 400) = 10/14 = 5/7.

If the question actually meant atleast 1/4. then it can be any number greater than 1000. so if the number of single marked fishes is 1/2 the answer wud be 2000/24000 = 5/6.

you can have effectively 1000 to 3600 as the number of single marked fishes. and each of these will give you a different probability.
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 [#permalink] New post 19 Dec 2004, 12:41
unless theres a catch to this question or im not reading it right

1/4 marked with one mark = 1/4 of 4000 = 1000 fishes

400 fishes with 2 or more marks

so total marked fishes = 1400 of which reqd prob is 1000/1400 = 10/14

but i doubt the question is that easy....

what od u guys think ....71.5 percent or 0.715
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 [#permalink] New post 19 Dec 2004, 16:02
Sargataur wrote:
unless theres a catch to this question or im not reading it right

1/4 marked with one mark = 1/4 of 4000 = 1000 fishes

400 fishes with 2 or more marks

so total marked fishes = 1400 of which reqd prob is 1000/1400 = 10/14

but i doubt the question is that easy....

what od u guys think ....71.5 percent or 0.715


target780, Sargataur,

Yes, the word 'at least 1/4' is part of the question stem [I am dead sure about this]. And No the answer is NOT 5/7 since this was "not" one of the five answer choices.
I too believe having 'at least' makes the problem unbounded and therefore this makes a good tricky problem which the ETS loves to throw into the questions.

Can anyone else give this problem a try.
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 [#permalink] New post 21 Dec 2004, 13:33
spiderman_xx wrote:
Sargataur wrote:
unless theres a catch to this question or im not reading it right

1/4 marked with one mark = 1/4 of 4000 = 1000 fishes

400 fishes with 2 or more marks

so total marked fishes = 1400 of which reqd prob is 1000/1400 = 10/14

but i doubt the question is that easy....

what od u guys think ....71.5 percent or 0.715


target780, Sargataur,

Yes, the word 'at least 1/4' is part of the question stem [I am dead sure about this]. And No the answer is NOT 5/7 since this was "not" one of the five answer choices.
I too believe having 'at least' makes the problem unbounded and therefore this makes a good tricky problem which the ETS loves to throw into the questions.

Can anyone else give this problem a try.


If this were a data sufficiency question I would say it is solvable, but it seems to be impossible to solve by hand. Since the probability is the number of favorable outcomes/total outcomes, you could add up the probability with 1/4 marked, then keep adding 1 more fish till you have all of them, or start with 0 marked, then go up to 1/4 and subtract 1 from this answer. Maybe there is some kind of pattern that would make any of these calculations possible but I don't know.

Was this problem in the OG? If so then do you know what number?
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 [#permalink] New post 21 Dec 2004, 13:44
toddmartin wrote:
spiderman_xx wrote:
Sargataur wrote:
unless theres a catch to this question or im not reading it right

1/4 marked with one mark = 1/4 of 4000 = 1000 fishes

400 fishes with 2 or more marks

so total marked fishes = 1400 of which reqd prob is 1000/1400 = 10/14

but i doubt the question is that easy....

what od u guys think ....71.5 percent or 0.715


target780, Sargataur,

Yes, the word 'at least 1/4' is part of the question stem [I am dead sure about this]. And No the answer is NOT 5/7 since this was "not" one of the five answer choices.
I too believe having 'at least' makes the problem unbounded and therefore this makes a good tricky problem which the ETS loves to throw into the questions.

Can anyone else give this problem a try.


If this were a data sufficiency question I would say it is solvable, but it seems to be impossible to solve by hand. Since the probability is the number of favorable outcomes/total outcomes, you could add up the probability with 1/4 marked, then keep adding 1 more fish till you have all of them, or start with 0 marked, then go up to 1/4 and subtract 1 from this answer. Maybe there is some kind of pattern that would make any of these calculations possible but I don't know.

Was this problem in the OG? If so then do you know what number?


toddmartin,
This was one of the questions that I got on my actual GMAT two weeks back. Of-course, the only rationale sol. I could get was 5/7, which was not one of the answer choices (I paniced). Therefore, I guessed.
Now, after seeing all of the responses, I guess this may have been an experimental question?
  [#permalink] 21 Dec 2004, 13:44
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