Answer to Question #305749 in Statistics and Probability for gurpinder

Question

1.     Suppose we know that 70% of all Canadians have or had a pet in their household. A researcher takes a simple random sample of 20 Canadians. What is the probability that at least 9 people in the sample have never had a pet in their household? Explain how you got the number. (2)

Answer 1

Let $X=$ the number of people who have never had a pet in their household: $X\sim Bin(n, p).$

Given $n=20, q=0.7, p=0.3$

$$P(X\ge9)=1-P(X<9)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)-P(X=4)$$

$$-P(X=5)-P(X=6)-P(X=7)$$

$$-P(X=8)=1-\dbinom{20}{0}(0.3)^0(0.7)^{20}$$

$$-\dbinom{20}{1}(0.3)^1(0.7)^{19}-\dbinom{20}{2}(0.3)^2(0.7)^{18}$$

$$-\dbinom{20}{3}(0.3)^3(0.7)^{17}-\dbinom{20}{4}(0.3)^4(0.7)^{16}$$

$$-\dbinom{20}{5}(0.3)^5(0.7)^{15}-\dbinom{20}{6}(0.3)^6(0.7)^{14}$$

$$-\dbinom{20}{7}(0.3)^7(0.7)^{13}-\dbinom{20}{8}(0.3)^8(0.7)^{12}$$

$$\approx0.11333146288$$