Answer to Question #100035 in Statistics and Probability for Tannah

Question

A fifth (20%) of a food retailer’s customers purchase bread each day at the store. If eight customers are chosen at random, use the binomial distribution to calculate the probability that at least three of them buy bread at the store each day.

Answer 1

Let $X=$ the number of a food retailer’s customers purchase bread each day at the store.

$X\sim B(N, p)$

$$P(X=x)=\binom{n}{x}p^x (1-p)^{n-x}$$

Given that $p=0.2, n=8.$ Then

$$P(X\geq3)=1-P(X=0)-P(X=1)-P(X=2)=$$ $$=1-\binom{8}{0}0.2^0 (1-0.2)^{8-0}-\binom{8}{1}0.2^1 (1-0.2)^{8-1}-$$ $$-\binom{8}{2}0.2^2 (1-0.2)^{8-2}=1-(0.8)^8 -8\cdot0.2(0.8)^7-$$ $$-{8! \over 2!(8-2)!}\cdot0.2^2(0.8)^6=1-0.16777216-0.33554432-$$ $$-0.29360128=0.20308224$$

The probability that at least three of them buy bread at the store each day

is $0.20308224$