Answer to Question #328081 in Statistics and Probability for robot

Question

A supervisor at an electric bulb factory examines bulbs produced in the factory for defects. She usually finds that there are 14 defective bulbs in a week (7 days).

a. What is the probability that there are exactly 3 defective bulbs in one day?

b. What is the probability that there are less than 5 defects in a day?

c. Suppose the supervisor joins work, find the probability that 72 hours go by before she discovers her first defective bulb.

Answer 1

$\lambda =14/7=2$

a.$P(X=3)=\frac{2^3e^{-2}}{3!}=0.18$

b.

$\\P(X<5)=P(0)+P(1)+P(2)+P(3)+P(4)=\\=e^{-2}(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!})=e^{-2}(1+2+2+1.33+0.67)=5$

c. $\lambda=2\times 3=6$

$P(X=0)=\frac{6^0e^{-6}}{0!}=0.0025$