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

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

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

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

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

$=0.1073741824+0.268435456$

$+0.301989888+0.201326592$

$=0.8791261184$