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Suppose red and blue cars come from two independent Poisson processes with rates $\lambda = 4$/hr and $\mu = 6$/hr. If our system starts empty find probability of

(a) no arrivals in the next 10 minutes

(b) no arrivals in the next 10 minutes, given the last arrival was over 9 minutes ago

(c) no red cars arriving in the next 10 minutes, but at least one blue car

(d) red car comes before blue car

(e) exactly 10 cars in the next hour

(f) exactly 10 cars, at least 9 of which are blue

(g) exactly 10 cars, and 5 of them came in the last hour.


I guess I need help because I am new to this. Please let me know if my work is ok. Let X(t), Y(t) be the number of arrival of red and blue car by time t. I will use the facts that X(t), Y(t) are Poisson distributed but the inter-arrival times are exponential i.i.d. First I convert $4$/hr and $6$/hr to $0.067/$minute and $0.1$/minute respectively. So,

(a) my answer is

$$P(X(10)=0)P(Y(10)=0) = \frac{(0.067)^{0}e^{-0.067 * 10}}{0!} \cdot \frac{(0.1)^{0}e^{-0.1 * 10}}{0!} =0.188247066. $$

(b) My answer is same as (a) since of memory less.

(c) My answer is

$$P(X(10)=0)P(Y(10)\geq 1) = e^{-0.067\cdot 10} \cdot (1 - P(Y(10)=0))=e^{-0.067\cdot 10}(1 -0.367879441) = 0.323456109. $$

(d) Not sure

(e) I can do it with all eleven cases. But maybe there is a good way?

(f, g) not sure. please assist me

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Time lacks at the moment, but let me provide you a handsome setup.

Let $X_{1},X_{2}\dots$ be iid random variables with distribution $\mathsf{Exponential}\left(10\right)$.

Let $R_{1},R_{2}\dots$ be iid random variables with distribution $\mathsf{Bernoulli}\left(0.4\right)$.

Also let the sequences $(X_n)_n$ and $(R_n)_n$ be independent wrt to eachother.

Let $S_{n}:=\sum_{k=1}^{n}X_{k}$ for positive integers $n$.

Let $N_{r}\left(t\right)=\left|\left\{ n\in\mathbb{N}_+\mid R_{n}=1\wedge S_{n}\leq t\right\} \right|$ (the number of red cars arriving in interval $\left[0,t\right]$)

Let $N_{b}\left(t\right)=\left|\left\{ n\in\mathbb{N}_+\mid R_{n}=0\wedge S_{n}\leq t\right\} \right|$ (the number of blue cars arriving in interval $\left[0,t\right]$)

Let $N\left(t\right):=N_{r}\left(t\right)+N_{b}\left(t\right)$ (the number of all cars arriving in interval $\left[0,t\right]$)

Then $N\left(t\right)=\left|\left\{ n\in\mathbb{N}_+\mid S_{n}\leq t\right\} \right|$ can directly be recognized as a Poisson proces with rate $10$.

Further it can be proved that $N_{r}\left(t\right)$ and $N_{b}\left(t\right)$ are two independent Poisson processes with rates $4$ and $6$ respectively.

This setup makes the answers on d), e) and f) easy and also simplifies the answer on a).

Disadvantage in this context might be that it must be proved what is asserted above about $N_r(t)$ and $N_b(t)$.

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