Answer to Question #86308 in Statistics and Probability for Ajay

Question

A typical sheet of metal has on the average 2 defects per 5 m^2
. What is the probability that a 10 m^2 sheet of metal will have at least 3 defects?

Answer 1

Assume the Poisson distribution.

$$P(X = k) = e^{-\lambda}*\lambda^k/k!$$

Let X denote the number of defects in a 10 m^2 sheet of metal. Then since the unit area is a 5 m^2 sheet of metal, we have

$$\lambda = 2 * 10/5 = 4$$

$$P(X⩾3)=1-e^{-4}*(4^0/0!+4^1/1!+4^2/2!)$$ $$P(X⩾ 3)=1-13*e^{-4}=0.7619$$