Answer to Question #303922 in Statistics and Probability for Alisia

Question

Global insurance has found that 20% (1 in 5) of all insurance policies are surrendered (cashed in) before their maturity date. Assume that 10 policies are randomly selected from the policies data base

What is the probability that:

No more than 3 of these 10 insurance policies will have been surrendered before their maturity date? No more than 3 orders will be recieved?

Answer 1

Let $X=$ the numbers of insurance policies surrendered (cashed in) before their maturity date: $X\sim Bin (n, p).$

Given $n=10, p=0.2,q=0.8$

$$P(X\le3)=P(X=0)+P(X=1)$$

$$+P(X=2)+P(X=3)$$

$$=\dbinom{10}{0}(0.2)^0(0.8)^{10-0}+\dbinom{10}{1}(0.2)^1(0.8)^{10-1}$$

$$+\dbinom{10}{2}(0.2)^2(0.8)^{10-2}+\dbinom{10}{3}(0.2)^3(0.8)^{10-3}$$

$$=0.1073741824+0.268435456$$

$$+0.301989888+0.201326592$$

$$=0.8791261184$$