Stage: 4 Challenge Level:
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
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 $760$
And just to make sure...
then, we must add that ratio together:
Hey presto!!! There are $800$ fish altogether!
And from Stephen at Blatchington Mill
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.