Answer to Question #350505 in Statistics and Probability for DasDurdle

Question

According to a report from the research for Studying Health System Change, 20% of South Africans delay or go without medical care because of concerns about cost. Suppose that 8 individuals are randomly selected. (a) What is the probability that two individuals will delay or go without medical care? (2) (b) What is the probability that at most two individuals will delay or go without medical care? (3) (c) What is the probability that at least seven individuals will delay or go without medical care? (3)

Answer 1

Let $X=$ the number of individuals who will delay or go without medical care: $X\sim Bin(n, p).$

Given $n=8, p=0.2, q=1-p=0.8$

a)

$$P(X=2)=\dbinom{8}{2}(0.2)^{2}(0.8)^{8-2}$$ $$=0.29360128$$

b)

$$P(X\le2)=P(X=0)+P(X=1)+P(X=2)$$ $$=\dbinom{8}{0}(0.2)^{0}(0.8)^{8-0}+\dbinom{8}{1}(0.2)^{1}(0.8)^{8-1}$$

$$+\dbinom{8}{2}(0.2)^{2}(0.8)^{8-2}=0.16777216$$

$$+0.33554432+0.29360128$$

$$=0.79691776$$

c)

$$P(X\ge 7)=P(X=7)+P(X=8)$$

$$=\dbinom{8}{7}(0.2)^{7}(0.8)^{8-7}+\dbinom{8}{8}(0.2)^{8}(0.8)^{8-8}$$

$$=0.00008192+0.00000256$$

$$=0.00008448$$