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A marine biologist wanted to estimate the number of fish in a large lake. He threw a net and found $30$ fish in the net. He marked all these fish and released them into the lake. The next morning he again threw the net and this time caught $40$ fish, of which two were found to be marked. The (approximate) number of fish in the lake is:

1. $600$
2. $1200$
3. $68$
4. $800$
5. $120$

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Answer should be $600,$ Option A.

The problem given is equivalent to the problem in which an urn contains some number of white balls in it. We take $30$ balls out of it, mark them and put them back into the urn. Now, we randomly take $40$ balls out of the urn, $2$ of them are found to be marked. What is the approximate number of balls that were present in the urn initially?

Solution: Suppose the urn contained $X$ balls initially. Then
if we take $n$ ball out of urn, probably $n\times (30/X)$ balls will be marked out of $n$ balls.
Here, $n = 40.$
So, Probably $40*(30/X)$ out of $40$ balls will be marked.

But it is given that there are $2$ marked balls,
So, $2 = 40\times (30/X),$ which gives

$X = (40 \times 30)/2 = 600.$

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+1

Below link might help in understanding the solution here.

http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/peter9.html

Biologist, marked 30 fish

Now , Next day we get 2 fish marked among 40 fish

therefore 30 fish marked among 40/2 *30=600 fish
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I think it is also equivalent to saying that For every $\mathbf{40}$ fish we are getting $\mathbf 2$ marked fish.

So, in order to get $\mathbf{30}$ fish we need to take out $\mathbf{40}$ fish $\frac{30}{2}\;\text{times}$.

So answer$= 40 \times 15 = 600$