If the lake contains a total of 1000 fish with 100 of them marked, what is the probability that when you catch 10 fish at least one of them is marked

Question

Since 100 fish are marked out of 1000, the chance would be 0.1, so it would be X = Bin(1000, .01)? If want to calculate P(X>=1), where X is the number of fish, would I used to poisson distribution?

Any help would be appreciated

Answer 1

Go step-by-step. Calculate what's the probability of not catching a marked fish in $10$ tries. At first there are $1000$ fishes and $100$ are marked, so there's $\frac {900}{1000}$ you won't caught a marked fish. Now let's fish again. $999$ fishes and $899$ unmarked. Do this $10$ times and you'll get:

$$\frac{900}{1000} \times \frac{899}{999} \times \cdots \times \frac{891}{991} \approx 34.69 \%$$

So now just subtract that from $100 \%$ and there's $65.31 \%$ you'll get at least one marked fish in 10 tries.

Answer 2

This is hypergeometric. An urn (the lake) has 100 red balls (marked fish) in 1000. We sample 10. The random variable X is the hypergeometric. We seek $P(X \ge 1) = 1 - P(X = 0),$ where

P(X = 0) = C(100, 0)C(900,10)/C(1000,10) = 0.3469.

Here I have used C to denote a binomial coefficient.

In R, either 'choose(900,10)/choose(1000,10)' or 'dhyper(0, 100, 900, 10)' will return this result. In general for larger numbers, calculations of such probabilities can overflow a computer (or calculator) along the way. The second R statement is optimized to avoid such difficulties.

The Capture-Recapture method of estimating the number of fish in a lake is a two-step process. First, capture c fish at random and mark them (here c = 100). Second, after fish have time to randomly mix themselves, capture r fish at random (here r = 10), and note the number x of marked fish among the r.

Estimating the number T of fish in the lake involves equating the proportion of marked fish in the lake c/T = 100/T to the proportion of marked fish in the sample x/r = x/10, then solving for T. In general T is estimated by t = cr/x (here t = 100(10)/x). So if you caught x = 2 marked fish, you would estimate T as 1000/2 = 500. Here the values of c and r are too small for a good estimate.

This method does not work if x = 0. So there is a practical basis for asking the original question. Modified estimators such as t = (c+1)(r+1)/(x+1) are often mentioned as avoiding division by 0 and also giving somewhat better estimates of T.

The hypergeometric distribution can be used to find confidence intervals for the estimate of T. There is rich literature capture-recapture methods, which have been used in many ecological settings, but also to estimate certain human populations. Variations seek to adjust for 'fish' that immigrate, emigrate, die, or give birth between the capture and recapture phases. (If you search the web, try to stick to probability texts and articles in reputable journals, there is a lot of total nonsense on this topic.)