Question

Tagged order is 0.1 probability that there are less than three tagged order forms in the sample of four.

Answer 1

Let $X=$ the number of tagged order forms: $X\sim Bin(n, p).$

Given $n=4, p=0.1,q=1-p=1-0.1=0.9.$

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

$$=\dbinom{4}{0}(0.1)^0(0.9)^{4-0}+\dbinom{4}{1}(0.1)^1(0.9)^{4-1}$$

$$+\dbinom{4}{2}(0.1)^2(0.9)^{4-2}=0.9963$$