Question

An industrial company estimates that the probability of an order arriving on time from a supplier is 0.7. Suppose 10 orders are placed this week. What is the probability that at least 8 arrives on time?

Answer 1

This is a binomial distribution with n=10, p=0.7.

$P(X\ge8)=P(X=8)+P(X=9)+P(X=10)=$

$=C_{10}^80.7^80.3^2+C_{10}^90.7^90.3^1+C_{10}^{10}0.7^{10}0.3^0=$

$=0.2335+0.1211+0.0282=0.3828.$