Answer to Question #347269 in Statistics and Probability for NANA

Question

A shop assistant makes 3 entry errors per week. Find the Probability that in a week chosen at random she will make:

(i) Exactly 2 errors [2 marks]

(ii) At least 2 errors [3 marks]

(b) What is the probability that over a period of two weeks, the shop assistant in part (b) above will commit at least 3 errors?

Answer 1

a) $\lambda=3$

(i)

$$P(X=2)=\dfrac{e^{-3}(3)^2}{2!}=0.224042$$

(ii)

$$P(X\ge2)=1-P(X=0)-P(X=1)$$ $$=1-\dfrac{e^{-3}(3)^0}{0!}-\dfrac{e^{-3}(3)^1}{1!}=0.800852$$

b) $\lambda t=3(2)=6$

$$P(X\ge3)=1-P(X=0)$$

$$-P(X=1)-P(X=2)$$

$$=1-\dfrac{e^{-6}(6)^0}{0!}-\dfrac{e^{-6}(6)^1}{1!}-\dfrac{e^{-6}(6)^2}{2!}$$

$$=0.938031$$