Nine out of every 100 people in the Philippines have Type B positive blood. If 8 people are chosen at random, what is the probability that less than 3 people have Type B blood?

Let $X=$ the number of people in the Philippines with Type B positive blood: $X\sim Bin (n, p).$

Given $n=8, p=0.09, q=1-p=1-0.09=0.91$

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

$=\dbinom{8}{0}(0.09)^0(0.91)^{8-0}+\dbinom{8}{1}(0.09)^1(0.91)^{8-1}$

$+\dbinom{8}{2}(0.09)^2(0.91)^{8-2}=0.97111328$