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spiderman_xx
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A fisherman caught 4000 fish. He marked at least 1/4 of the 19 Dec 2004, 10:52
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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?

I do not have the answer choices. Please explain your answer.



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target780
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A fisherman caught 4000 fish. He marked at least 1/4 of the 19 Dec 2004, 12:40
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Quote:
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?
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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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Sargataur
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A fisherman caught 4000 fish. He marked at least 1/4 of the 19 Dec 2004, 12:41
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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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A fisherman caught 4000 fish. He marked at least 1/4 of the 19 Dec 2004, 16:02
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Sargataur
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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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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toddmartin
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A fisherman caught 4000 fish. He marked at least 1/4 of the 21 Dec 2004, 13:33
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spiderman_xx
Sargataur
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.
Show more


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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spiderman_xx
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A fisherman caught 4000 fish. He marked at least 1/4 of the 21 Dec 2004, 13:44
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toddmartin
spiderman_xx
Sargataur
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?
Show more


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?



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!