Answer to Question #170125 in Statistics and Probability for bilawal haq

Question

A small lake contains 25 fish. One day a fisherman catches 8 of these fish and tags them so that they can be recognized if they are caught again. The tagged fish are released back into the lake. The next day the fisherman goes out and catches 5 fish, which are kept in the fishing boat until they are all released at the end of the day. • What is the distribution of the number of tagged fish caught by the fisherman on the second day? • Find the expected number of tagged fish recaptured on the second day. • Find the probability that the fisherman caught at least 6 tagged fish on the second day

Answer 1

We have that:

8 of 25 fishes are tagged, thus p = 8/25 = 0.32

n = 5

This follows the binomial distribution

The expected number of fish: p * n = 5 * 0.32 = 1.6

The binomial probability is calculated by the formula:

$P(X=m)=C(n,m)\cdot p^m\cdot(1-p)^{n-m}$

$P(X\ge6)=1-P(X<6)$

$P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$

It is not possible to calculate the probability that the fisherman caught at least 6 tagged fish on the second day from 5 trials since n must be greater than or equal to m ($n\ge m)$

The general solution is listed above.