Question

5. A certain electronics company produces a particular type of vacuum tube. On the average, 4 out of 100 are defective. The company packs the tubes in boxes of 400. Let 16. Find the probability that a box of 400 tubes will contain: (This is a Poisson Distribution) a) exactly 3 defectives b) at least 3 defective tubes c) at most 3 defective

Answer 1

a)

P(X=3) =e^-16*16³/3! =0.000077

b)

P(X>=3) =1-P(X<=2) =1-(P(X=0)+P(X=1)+P(X=2)) =1-(e^-16*16⁰/0!+e^-16*16¹/1!+e^-16*16²/2!)=1-0.000016 =0.999984

c)

P(X<=3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)=e^-16*16⁰/0!+e^-16*16¹/1!+e^-16*16²/2!+e^-16*16³/3! =0.000093