## Statistics for the Life Sciences (5th Edition)

P(X=j)$\frac{n!}{j!(n-!)}$$p^{j}$$(1-p)^{n-j}$
a.) P(none has high blood lead) = $\frac{16!}{0!16!}$$0.125^{0}$$(1-0.125)^{16}$ = 0.1181 b.) P(1 has high blood lead) = $\frac{16!}{1!15!}$$0.125^{1}$$(1-0.125)^{15}$ = 0.2699 c.) P(2 have high blood lead) = $\frac{16!}{12!14!}$$0.125^{2}$$(1-0.125)^{14}$ = 0.2891 d.) P(3 or more have high blood lead) = 1 - P(none)+P(1)+P(2) P(3 or more) = 1 - [(0.1181) + (0.2699) + (0.2891)] = 0.3229