I'm just curious if my work is correct, and if not, where I made a mistake.

My Task:

Cars arrive according to a Poisson process with a rate of 12 per hour.
(1) What is the probability that the third car that arrives after 8 AM arrives within 10 minutes of the first car that arrives after 8 AM?
(2) A car has a $1/4$ chance of being Japanese. What is the probability that exactly 5 Japanese and 10 non-Japanese cars arrive between 8 and 9 AM?

My answer:

(1) The question is equivalent to asking for the probability of three cars arriving in a 10-minute interval. It follows from the rate of 12 per hour that, on average, 2 cars arrive every 10 minutes. Then we just use the Poisson formula with $\lambda = 2$: $$P(X=3) = \frac{\lambda^3}{3!}e^{-2} = \frac{8}{6}e^{-2} \approx 0.18$$ (The above is wrong -- see Heropup's answer and my comment below.)

(2) Note that the arrival of the car is independent of its make. So we multiply the probability that 15 cars arrive in the interval (Poisson, X) by the probability that of the 15 cars, 10 are non-Japanese and 5 are Japanese (Binomial, Y). For Poisson, we let $\lambda = 12$. Thus we get: $$P(X = 15) \cdot P(Y=5) = \frac{12^{15}}{15!}e^{-12} \cdot {15 \choose 5}(1/4)^{5}(3/4)^{10} \approx 0.012 $$

  • $\begingroup$ Is the second part correct? From where have you picked up this problem? I would like to practice binomial and poisson sums, so any suggestions are welcome! $\endgroup$ – user585380 Sep 9 '18 at 9:52

Your reasoning for (1) is incorrect. Because the process is Poisson, interarrival times are independent. So after the first car has arrived, subsequent arrivals occur as they would as if the process had just begun. But then, the third car to arrive is the second car we observe after the arrival of the first car. In other words, if the cars have arrival times $T_1$, $T_2$, $T_3$, then $T_3$ corresponds to the second arrival time after $T_1$. Thus, we must compute the probability that $T_3 - T_1 \le 10$, or equivalently, that at least two additional cars are observed in the 10 minutes following the observation of the first car.

  • $\begingroup$ Ahh, I see. So I can take the arrival of the first car as a given, and from that point want to find the probability that at least two cars arrive within 10 minutes? Then I want to compute $P(X \geq 2) = 1 - (P(X=1) + P(X=0))$, which I can compute with the Poisson formula? EDIT: I have computed this to be $P(X \geq 2 ) = 1 - (3e^{-2}) \approx 0.41 = 0.59$. $\endgroup$ – Newb May 22 '15 at 9:17
  • $\begingroup$ And (2) looks ok? $\endgroup$ – Newb May 22 '15 at 9:19

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