Question 1. [9 marks]. Arrivals of the Number 1 bus at a particular bus stop form aPoission process with rate parameter λ1 and arrivals of the Number 2 bus at the samebus stop form a Poisson process with rate parameter λ2. The two Poisson processes areindependent.(a) What is the probability that precisely 4 buses (of type Number 1 and/or Number 2)arrive in the time interval [0, t]?(b) What is the probability that exactly 3 Number 2 buses pass while I am waiting for aNumber 1 bus?(c) When the maintenance depot goes on strike half the buses break down before they reachmy stop. What then is the probability that not a single bus passes in the time interval[0, t].When answering each of parts (a), (b) and (c) you should briefly explain your reasoning.Hints: (i) Recall that ∫ ∞0xα−1e−βxdx = Γ(α)βα ,where the gamma function Γ(·) is such that, for a positive integer n, Γ(n) = (n − 1)!(ii) In part (c) you may assume that thinning has taken place, as defined in the lecture notes
Question
Question 1. [9 marks]. Arrivals of the Number 1 bus at a particular bus stop form aPoission process with rate parameter λ1 and arrivals of the Number 2 bus at the samebus stop form a Poisson process with rate parameter λ2. The two Poisson processes areindependent.(a) What is the probability that precisely 4 buses (of type Number 1 and/or Number 2)arrive in the time interval [0, t]?(b) What is the probability that exactly 3 Number 2 buses pass while I am waiting for aNumber 1 bus?(c) When the maintenance depot goes on strike half the buses break down before they reachmy stop. What then is the probability that not a single bus passes in the time interval[0, t].When answering each of parts (a), (b) and (c) you should briefly explain your reasoning.Hints: (i) Recall that ∫ ∞0xα−1e−βxdx = Γ(α)βα ,where the gamma function Γ(·) is such that, for a positive integer n, Γ(n) = (n − 1)!(ii) In part (c) you may assume that thinning has taken place, as defined in the lecture notes
Solution
(a) Since the two Poisson processes are independent, we can add them together to form a new Poisson process with rate parameter λ = λ1 + λ2. The probability that exactly 4 buses arrive in the time interval [0, t] is given by the Poisson probability mass function:
P(X = 4) = (e^(-λt) * (λt)^4) / 4!
where X is the random variable representing the number of buses that arrive in the time interval [0, t].
(b) The probability that exactly 3 Number 2 buses pass while waiting for a Number 1 bus is given by the Poisson probability mass function:
P(Y = 3) = (e^(-λ2t) * (λ2t)^3) / 3!
where Y is the random variable representing the number of Number 2 buses that pass in the time interval [0, t]. Note that the arrival of Number 1 buses does not affect this probability because the two processes are independent.
(c) When the maintenance depot goes on strike, half the buses break down before they reach the stop. This can be modeled as a thinning of the Poisson process, where each bus independently has a 0.5 probability of breaking down. The new rate parameters are λ1' = λ1 / 2 and λ2' = λ2 / 2. The probability that no buses pass in the time interval [0, t] is given by the Poisson probability mass function:
P(Z = 0) = e^(-(λ1' + λ2')t)
where Z is the random variable representing the number of buses (of either type) that pass in the time interval [0, t].
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