A poultry farmer knws that frm past experience that 5% of day old chicks are infected with a disease.He selected 4 chicks at random for inspection.

Question

Find the probability that he will find

a..None infected

b.one infected

c.at most one infected

d.at least one infected

Answer 1

The probability of non-infection is 95% or 0.95.

a) The probability of all 4 chicks being uninfected is 0.95^4=0.8145 approx=81.45%.

b) Suppose the first chick is infected and the remaining three are not, probability=0.05*0.95^3=0.0429 approx=4.29%. But it could have been any one of the four that was infected so we need to multiply by 4=0.1715 or 17.15%.

c) At most one implies the sum of the probabilities of none infected and one infected, which is (a)+(b)=98.60%.

d) This is the same as 1-(a), because the complement of none infected means at least one is infected. So 1-0.8145=0.1855 or 18.55%.

Comments

Thanks