Question

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate α = 8 per hour. Suppose that with probability 0.5 an arriving vehicle will have no equipment violations.
(a) What is the probability that exactly eight arrive during the hour and all eight have no violations? (Round your answer to four decimal places.)

(b) For any fixed y ≥ 8, what is the probability that y arrive during the hour, of which eight have no violations?

(c) What is the probability that eight "no-violation" cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from y = 8 to [infinity].] (Round your answer to three decimal places.)

Answer 1

Answer:

a) The probability that exactly eight arrive during the hour and all eight have no violations is 0.0005.

b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:

$P(X=y\,\&\,nv=8)=\frac{4^ye^{-8}}{8!(y-8)!}$

c) The probability that eight "no-violation" cars arrive during the next hour is 0.030.

Step-by-step explanation:

a) The probability that exactly eight arrive during the hour and all eight have no violations is equal to the product of the probability of arrival of 8 vehicules and the probability of having 8 vehicules with no violations.

$P(X=8\,\&\,no\, violations)=P(no\, violations|X=8)*P(X=8)\\\\P(X=8\,\&\,nv)=(0.5)^8*\frac{8^8e^{-8}}{8!} = 0.0039*0.1396=0.0005$

b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:

$P(X=y\,\&\,nv=8)=P(nv=8|X=y)*P(X=y)\\\\P(X=y\,\&\,nv=8)=[\binom{y}{8}(0.5)^8*(0.5)^{y-8}]*\frac{8^ye^{-8}}{y!} =\frac{y!}{8!(y-8)!}0.5^y *8^y*\frac{e^{-8}}{y!}\\\\ P(X=y\,\&\,nv=8)=(\frac{y!}{y!})(0.5*8)^y\frac{e^{-8}}{8!(y-8)!}=\frac{4^ye^{-8}}{8!(y-8)!}$

c) Using the result of point (b) we can express the probability that eight "no violation" vehicules arrive durting the next hour as:

$P(nv=8)=\sum\limits^\infty_{y=8} {\frac{4^ye^{-8}}{8!(y-8)!}}=\frac{e^{-8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^y}{(y-8)!}}=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^{y-8}}{(y-8)!}}\\\\P(nv=8)=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{z=0} {\frac{4^{z}}{z!}}=\frac{e^{-8}4^{8}}{8!}*e^4=\frac{e^{-4}4^{8}}{8!}\\\\P(nv=8)= 0.030$