Assembled units arrive at an inspection station following a Poisson process with

Question

Assembled units arrive at an inspection station following a Poisson process with rate lambda.

The units are inspected upon arrival, and a portion p 2 (0; 1) of the units is identied

as defective. (Assume inspection time is negligible.) Suppose at t = 0, a defective unit is

identied (upon arrival). Let T denote the time until the next defective unit is identied.

(i) What distribution does T follow?

(ii) What is the probability that given two units have arrived in [0; T] and both are good?

(iii) What is the probability that the second good unit arrives before the second defective unit?

Explanation / Answer

Solution

Back-up Theory

“a portion p of the units is identified as defective.” =>

probability of finding a defective is p………………………………………………………………………(1)

(1) => probability of finding a good unit is (1 - p)………………………..................................………(2)

If X = Number of events per unit time, Y = inter-event time, and X ~ Poisson (), then

Y ~ Exponential(), where = 1/ . ……………………………………..............................……………(3)    

Now, to work out the solution,

Part (a)

Given,“Assembled units arrive at an inspection station following a Poisson process with rate lambda.”

=> by (3) of Back-up Theory, T = the time until the next defective unit is identified has Exponential(),

where = 1/ ANSWER

Part (b)

Probability of finding both units good = (1- p)2 [by (2) of Back-up Theory] ANSWER

Part (c)

probability that the second good unit arrives before the second defective unit =>out of four units arrived, first two are good and defective in either order, third unit is good and the fourth unit is defective.

This probability = p(1 - p)(1 - p)p + (1 - p)p(1 - p)p [first is probability of defective, good, good, defective and second is good, defective, good, defective]

= 2p2(1 - p)2 ANSWER

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