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A worker for the Department of Fish and Game is assigned the job of estimating the number of trout in a certain lake of modest size. She proceeds as follows: She catches 100 trout, tags each of them, and puts them back in the lake. One month later, she catches 100 more trout, and notes that 10 of them have tags.

(a) Without doing any fancy calculations, give a rough estimate of the number of trout in the lake.

There are 100 tagged trout, and tagged trout represent one-tenth of the trout caught, so we can estimate the number of trout in the lake at 1,000.

(b) Let N be the number of trout in the lake. Find an expression, in terms of N, for the probability that the worker would catch 10 tagged trout out of the 100 trout that she caught the second time.

This situation can be modeled with a hypergeometric probability distribution. There are N fish, 100 of them are tagged, and we want the probability of catching 100 fish, including 10 of those 100 tagged fish. That probability would be found with this expression:

\(\frac{\binom{100}{10}\binom{N-100}{90}}{\binom{N}{100}}\)

This expression can be rewritten as:

\(\frac{\binom{100}{10}\binom{N-100}{90}}{\binom{N}{100}}=\frac{\frac{100!}{90!10!}\frac{(N-100)!}{90!(N-190)!}}{\frac{N!}{100!(N-100)!}}=\frac{(100!)(100!)((N-100)!)((N-100)!)}{(90!)(90!)(10!)(N!)((N-190)!)}\)

(c) Find the value of N which maximizes the expression in part (b). This value is called the maximum likelihood estimate for the unknown quantity N . Hint: Consider the ratio of the expressions for successive values of N.

Define \(f(N)=\frac{(100!)(100!)((N-100)!)((N-100)!)}{(90!)(90!)(10!)(N!)((N-190)!)}\)

Then \(f(N+1)=\frac{(100!)(100!)((N-99)!)((N-99)!)}{(90!)(90!)(10!)((N+1)!)((N-189)!)}=f(N)\times \frac{(N-99)(N-99)}{(N+1)(N-189)}=f(N)\times \frac{N^2-198N+9801}{N^2-188N-189}\)

Now define \(g(N)=\frac{N^2-198N+9801}{N^2-188N-189}\). Therefore, \(f(N+1)=f(N)\times g(N)\). This means that as long as \(g(N)>1\), increasing the number of fish by 1 will increase the value of \(f(N)\) (and therefore the number of fish in the pond does not maximize the expression). We can find the maximum by finding the value of \(N\) which will result in \(g(N)=1\), after which the value of \(g(N)\) will drop below 1 and further increases in the number of fish in the pond will decrease the value of the expression.

In order to get \(g(N)=1\), it must be true that \(N^2-198N+9801=N^2-188N-189\), or \(10N=10000\), resulting in \(N=1000\). Therefore, at 1000 fish in the pond the value of the expression is maximized.