Question

Suppose we have a population of N deer in a study area. Initially n deer from this population are captured, marked so that they can be identified as having been captured, and returned to the population. After the deer are allowed to mix together, m deer are captured from the population and the number k of these deer having marks from the first capture is observed. Assuming that the first and second captures can be considered random selections from the population and that no deer have either entered or left the study area during the sampling period, what is the probability of observing k marked deer in the second sample of m deer

Answer 1

`P(k) = ((n C k) [(N-n) C (m-k)])/((NCm))`

Explanation:

Step 1: Number of possible combination of selecting ‘m’ deer in second sample

Total number of deer are N and therefore the combinations can be calculated as (N С m).

Step 2: Number of possible combination of marked deer ‘k’ in second sample

Total number of marked deer in total population is ‘n’. Therefore, the possible number of selecting marked deer is (n C k).

Step 3: Number of possible combination unmarked deer in second sample

Since we have already calculated the total combinations of selecting marked deer in the second sample. Hence, we have to calculate the total unmarked deer in total population which is N-n and number of unmarked deer in the second sample which is m-k.

Therefore, total possible combination of unmarked deer in second sample is [(N-n) C (m-k)].

Step 4: Probability of selecting unmarked deer in the second sample is

Let the probability of selecting unmarked deer in the second sample be P(k)

Therefore,

`P(k) = ((n C k) [(N-n) C (m-k)])/((NCm))`