The assumption that the fraction of marked individuals found in a second sample mirrors the fraction marked in the whole population:

where $M$ = marked in the first sample, $C$ = size of the second sample, $R$ = marked ones recaptured in it, and $N$ = the (unknown) total population you're estimating.

Substitute the three known values and cross-multiply to solve for $N$.

90 birds ringed ($M=90$); second sample of 120 ($C=120$) includes 20 ringed ($R=20$):

Four specific ideas are examinable, not vague ones like "it's just an estimate":

• No mark loss — tags/dye didn't fall off or fade.
• Random mixing — the marked individuals spread back out evenly, so the second sample is representative.
• Closed population — no births, deaths, immigration or emigration between samples.
• No change to survival/behaviour — the mark doesn't make the animal easier or harder to recatch.

If marks are lost, $R$ comes out artificially low, which inflates $N$ (since $N=\frac{M\times C}{R}$ — a smaller $R$ in the denominator makes $N$ bigger). Being able to explain this direction of error, not just state the assumption, is what separates a Grade 7 answer from a Grade 6 one.

The same proportion rearranges to solve for any of the four variables — just substitute what's unknown as $X$ and solve.

$C=50$, $R=32$, $N=125$, find $M$:

Grade 9 variants replace numbers with expressions, usually linked by a second piece of information (like "$T$ is one-fifth of $N$"). Substitute everything into the same proportion and solve as normal.

A scientist tags $T$ lions; the next day finds $T-10$ lions, 4 of which are tagged; $T$ is $\frac15$ of the total $N$ (so $N=5T$). Find $N$.

1. $\dfrac{R}{C}=\dfrac{M}{N} \Rightarrow \dfrac{4}{T-10}=\dfrac{T}{5T}=\dfrac15$.
2. Cross-multiply: $4\times5 = T-10 \Rightarrow 20=T-10 \Rightarrow T=30$.
3. $N=5T=150$.

A park ranger marks 400 fish and releases them. The next day she catches 60 fish, of which 3 are marked. Estimate the total fish population, and state one assumption your estimate relies on.

1. $\dfrac{3}{60}=\dfrac{400}{N} \Rightarrow 3N = 400\times60=24000 \Rightarrow N=8000$.
2. Assumption: the marked fish mixed back thoroughly and randomly with the rest of the population before the second sample was taken (any one specific, named assumption from the list above is acceptable).

Population estimates must always round to a sensible whole number — a population of "9333.3 jelly beans" should be rounded, since individuals can't be fractional.