Answer:
0.8187,0.00115,0.20
Explanation:
Given that one out of 5000 individuals in a population carries a certain defective gene.
Hence in the sample of 1000, we find X no of those carrying defective gene will follow a Poisson distribution with mean

a)
the probability that none of the sample individuals carries the gene

b) the probability that more than two of the sample individuals carry the gene
=

c) the mean of the number of sample individuals that carry the gene
0.20