Bily wonted to estimate the number of fish in a lake.

She randamly caplured Se fist, then lagged and released them

One week later she randemly captured 10 Fish From the same lake. She found that 12 of these 40 Rish were tagged.

What is the best estimate for the tolal number of fish in the lake?

A 58
B 70
C. 82
D. 100

Question

Bily wonted to estimate the number of fish in a lake.

She randamly caplured Se fist, then lagged and released them

One week later she randemly captured 10 Fish From the same lake. She found that 12 of these 40 Rish were tagged.

What is the best estimate for the tolal number of fish in the lake?

A 58
B 70
C. 82
D. 100

UpStudy AI · Accepted Answer

We use the mark–recapture method. Let  
$$
N
$$
be the total number of fish in the lake.

In a typical mark–recapture experiment the fraction of tagged fish in the second sample is assumed to be the same as the fraction of tagged fish in the whole lake. That is,
$$
\frac{\text{number tagged in second sample}}{\text{size of second sample}} = \frac{\text{number tagged in initial sample}}{N}.
$$

In this problem the numbers (after correcting the misprints) are interpreted as follows:
- In the first capture, Bily tagged $$30$$ fish.
- In the second capture, she caught $$40$$ fish, of which $$12$$ were tagged.

So we have
$$
\frac{12}{40} = \frac{30}{N}.
$$

Now solve for $$N$$:
1. Cross multiply:
$$
12N = 30 \times 40.
$$
2. Compute the product on the right:
$$
12N = 1200.
$$
3. Divide both sides by 12:
$$
N = \frac{1200}{12} = 100.
$$

Thus, the best estimate for the total number of fish in the lake is $$\boxed{100}$$.
The best estimate for the total number of fish in the lake is 100.

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