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 ∞.] (Round your answer to three decimal places.)

Answer 1

a)

Find the probability that exactly eight arrive during the hour and all eight have no violations.

$$ \begin{aligned} P(X=8 \text { and no violation }) &=P(\text { no violation } \mid X=8) P(X=8) \\ &=\frac{e^{-\alpha t} \alpha t^{x}}{x !} P(X=8) \\ &=\frac{e^{-8 x 1} 8 \times 1^{8}}{8 !}(0.5)^{8} \\ &=\frac{e^{-8} 8^{8}}{8 !}(0.5)^{8} \\ &=0.00390625(0.1395865) \\ &=0.00054 \end{aligned} $$

Hence, the required probability is \(0.00054\).

b)

Find the probability that \(y\) arrive during the hour, of which eight have no violations.

$$ \begin{aligned} P((Y=y) \cap(8 \text { no violations })) &=P(8 \text { no violation } \mid Y=y) P(Y=y) \\ &=\left[\left(\begin{array}{l} y \\ 8 \end{array}\right)(0.5)^{8}(1-0.5)^{y-8}\right] \frac{e^{-8} 8^{y}}{y !} \\ &=\left[\frac{y !}{(y-8) ! 8 !}(0.5)^{y}\right] \frac{e^{-8} 8^{y}}{y !} \\ &=\frac{e^{-8} 4^{y}}{8 !(y-8) !} \end{aligned} $$

c)

Find the probability that eight "no-violation" cars arrive during the next hour.

$$ \begin{aligned} P(8 \text { no violations }) &=\sum_{y=8}^{\infty} \frac{e^{-8} 4^{y}}{(y-8) ! 8 !} \\ &=\frac{e^{-8}}{8 !} \sum_{y=8}^{\infty} \frac{4^{y}}{(y-8) !} \\ &=\frac{e^{-8}}{8 !}\left[\frac{4^{8}}{1}+\frac{4^{9}}{1 !}+\frac{4^{10}}{2 !}+\ldots . .\right] \\ &=\frac{e^{-8} 4^{8}}{8 !}\left[1+\frac{4^{9}}{1 !}+\frac{4^{10}}{2 !}+\ldots .\right] \\ &=\frac{e^{-8} 4^{8}}{8 !} e^{4} \\ &=0.02977 \end{aligned} $$

Hence, the required probability is \(0.02977\).