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Weekly Problem 4 - 2012
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Counting Fish

Stage: 4 Challenge Level: Challenge Level:1

Here's a couple of nice explanations :

From Francesca at Wimbledon High School : by the end of that week, the $40$ marked fish would have spread out and integrated with the rest of the fish population. So that the second lot of $40$ fish that are caught would be a thoroughly mixed group of marked and unmarked fish.

Count the number of marked and unmarked fish, get a ratio, then reason like this :

For example if in the second lot of $40$ fish, the marked to unmarked ratio was $2 : 38$ and we know that there are $40$ marked fish altogether, we might assume that the ratio in the second lot of fish is close to the ratio for all the fish together.

So $2 : 38$ matches $40 : ?$
$40/2$ is $20$ and then scaling that up by $38$ to get $760$ ($20\times38$)
And just to make sure...
$$38/2=19 $$
$$760/40=19 $$
then, we must add that ratio together:
Hey presto!!! There are $800$ fish altogether!

Thanks Francesca.

And from Stephen at Blatchington Mill School :

If you mark $40$ fish and release them. Then, when those fish have had time to mix back in with the whole population, catch another $40$. If you count the number ($c$) in that second catch which are already marked, you can estimate the size of the whole population ($T$) by saying that $40 = T(c/40) $

For example if when you looked at the second catch and $10$ were already marked then you could estimate that you had marked a quarter of the entire population of fish, because that's the proportion in the sample, so the original $40$ were about one quarter of the whole population and there are therefore approximately $120$ fish.