Question

Suppose that minor defects are distributed over the length of a cable according to a Poisson process with rate 9 per meter and, independently, major defects are distributed over the same cable according to a Poisson process with rate 1 per meter. a. If exactly 1 minor defect has been found in the first 1 meter of the cable, what is the probability it is within the first 20 cm b. If exactly 2 defects have been found in the first 1 meter of the cable, what is the Probability that one is a minor and other is a major defect?

Answer 1

a) expected number of minor defects in 20 cm =(20*9/100)=9/5

and expected number of minor defects in 80 cm =(80*9/100)=36/5

therefore P(1 minor defect within 20 cm given 1 minor defect in 1meter)

=P(1 minor defect in 20 cm)*P(0 minor defect in next 80 cm)/P(1 minor defect in 1 meter)

=(e^-9/5*(9/5)¹/1!)*(e^-36/5*(36/5)⁰/0!)/(e^-9*(9)¹/1!) =0.2

b)

let x1 and x2 are minor and major defects

total number of defects in 1 meter =X =x1+x2

expected number of defects in 1meter E(x)=9+1 =10

P( one is a minor and other is a major defect given total 2 defects in 1 meter)

=P(X1=1)*P(X2=1)/P(X=2) =(e^-9*9¹/1!)*(e^-1*1¹/1!)/(e^-10*10²/2!) =0.18