Automobiles arrive at a vehicle equipment inspection station according to a pass

Question

Automobiles arrive at a vehicle equipment inspection station according to a passion process with rate a = 10 per hour. Suppose that with probability 0.5 an arriving vehicle will have no equipment violations. (a) What is the probability that exactly ten arrive during the hour and all ten have no violations? (Round your answer to four decimal places.) (b) For any fixed y greaterthanorequalto 10, what is the probability that y arrive during the hour, of which ten have no violations? (c) What is the probability that ten 'no-violation' cars arrive during the next hour? You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

Solution

Let X = number of automobiles arriving in an hour. Then, we are given X ~ Poisson(10)

Let Z = number of ‘no violation’ automobiles out of n automobiles arriving in an hour.

Then, Z ~ B(n, ½) since we are given that probability of ‘no violation’ is 0.5.

Back-up Theory

1. Probability mass function (pmf) of X is given by

P(X = x) = e – .x/(x!), x = 0, 1, 2, ……. , . …………………………………….(1)

2. Probability mass function (pmf) of Z is given by

p(x) = P(X = x) = (nCx)(½)n, x = 0, 1, 2, ……. , n ………………………………..(2)

Values of p(x) for various values of (, x) and (n, p, x)can be obtained by using Excel Function

Part (a)

Probability of 10 arrivals in an hour and all 10 having ‘no violation’ = P(X= 10, Z = 10)

= P(X = 10) x P(Z = 10) = {e – 10.1010/(10!)}{ (10C10)(½)10 }

= 0.12511 x 0.000977 = 0.000122178 = 0.0001 ANSWER

Part (b)

For a fixed y 10, probability of y arrivals and 10 out of y have ‘no violation’

= P(X= y, Z = 10) = P(X = y) x P(Z = 10)

= {e – 10.10y/(y!)}{ (yC10)(½)10 }, y = 10, 11, … ANSWER

Part (c)

Probability of 10 ‘no violation’ arrive in the next hour = probability of y arrivals in the next hour and out of these y, 10 are ‘no violation’. Since number of ‘no violation’ is 10, y 10.

Thus, the required probability = sum of the probability in Part (b) over all values of y 10

= [10, ]{e – 10.10y/(y!)}{ (yC10)(½)10 }

= (e – 10)(½)10[10, ]{(yC10)(10y)/(y!)}ANSWER

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