EXAM1SOL.pdf - Past Exam 1 Solutions MTH3241 MTH3241 Past Exam 1 Solutions Page 1 of 11 Past Exam 1 MTH3241 Solutions 1 Suppose that the number of calls

EXAM1SOL.pdf - Past Exam 1 Solutions MTH3241 MTH3241 Past...

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Past Exam 1 Solutions MTH3241 MTH3241 Past Exam 1 Solutions Page 1 of 11
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Past Exam 1 Solutions MTH3241 1 Suppose that the number of calls per hour arriving at an answering service follows a Poisson process with parameter λ . (a) What is the probability that fewer than two calls (i.e. at most one call) come in the first hour? Let X t be the number of calls by time t . X t is Poisson λt . P (at most 1 call in 1st hour) = P ( X 1 1) = P ( X 1 = 0) + P ( X 1 = 1) = e λ + λe λ = (1 + λ ) e λ . 2 (b) Suppose that six calls arrive in the first hour. What is the probability that at least two calls arrive in the second hour? Since X 2 X 1 is Poisson λ and is independent of X 1 , P (at least 2 calls in 2nd hour | 6 calls in 1st hour) = P ( X 2 X 1 2 | X 1 = 6) = P ( X 1 2) = 1 P ( X 1) = 1 (1 + λ ) e λ . 2 (c) The person answering the phone waits until fifteen phone calls have arrived before going to lunch. What is the expected amount of time that the person will wait? T 15 , the time to the fifteenth arrival is Erlang of order 15 and parameter λ . Thus E [ T 15 ] = 15 λ . 2 (d) Suppose it is known that exactly eight calls arrived in the first two hours. What is the probability that exactly five of them arrived in the first hour? Applying Bayes’ rule and the facts that X 2 X 1 is Poisson λ and is inde- pendent of X 1 and that X t is Poisson λt we get, P ( X 1 = 5 | X 2 = 8) = P ( X 2 = 8 | X 1 = 5) P ( X 1 = 5) P ( X 2 = 8) = P ( X 1 = 3) P ( X 1 = 5) P ( X 2 = 8) = e λ λ 3 3! e λ λ 5 5! e 2 λ (2 λ ) 8 8! = ( 8 3 ) 1 2 8 . 3 (e) Suppose it is known that exactly k calls arrived in the first t hours. What is the probability that exactly j of them ( j k ) arrived in the first s hours ( s t )? Page 2 of 11
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Past Exam 1 Solutions MTH3241 Applying Bayes’ rule and the facts that X t X s is Poisson λ ( t s ) and is independent of X s and that X t is Poisson λt we get, P ( X s = j | X t = k ) = P ( X t = k | X s = j ) P ( X s = j ) P ( X t = k ) = P ( X t s = k j ) P ( X s = j ) P ( X t = k ) = e λ ( t s ) ( λ ( t s )) k - j ( k j )! e λs ( λs ) j j ! e λt ( λt ) k k ! = ( k j ) ( s t ) j ( 1 s t ) k j . 3 (f) Deduce the conditional mean number of arrivals in the first s hours given that exactly k calls arrived in the first t hours ( t s ). It follows from (e), that the conditional distribution of X s given that X t = k is binomial with parameters k and s/t . Thus E [ X s | X t = k ] = ks/t . 2 [14 marks] 2 (a) Consider the four-state Markov chain with transition matrix 1 0 0 0 p 0 q 0 0 p 0 q 0 0 0 1 where q = 1 p and 0 < p < 1. i. Represent graphically this Markov chain. 1 p q p q 1 1 2 3 4 2 ii. Assume that the chain starts in state 2. Find the absorption probabilities. Page 3 of 11
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Past Exam 1 Solutions MTH3241 The chain can be absorbed in either state 1 or state 4 – states 2 and 3 are transient. Let a i denote the probability of absorption into state 1, having started from state i . Then the absorption probability equations are: a 1 = 1 , a 4 = 0 , a 2 = pa 1 + qa 3 , a 3 = pa 2 + qa 4 .
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