Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and released to mix into the population. After they have had an opportunity to mix, a random sample of 10 of these animals is selected. Let X = the number of tagged animals in the second sample. Assuming there is a total of 25 animals of this type in the region, what are E(X) and Var(X)?

The hypergeometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

[tex]E(X)= n\frac{M}{N}[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}[/tex]

What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]

What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

[tex]P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565[/tex]

What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

[tex]P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474[/tex]