Answers to the exam ISP of December 19, 2005 1. Counting time in hours Summary: Answers to the exam ISP of December 19, 2005 1. Counting time in hours: a. Taking t = 2 in P(N(t) = 20) = e-10t (10t)2 0/20! yields 0.089 b. P(N(0.01) 1) = 1 - e-0.1 , or P(X 0.01) where X is an interarrival time, hence exponentially distributed with parameter 10. c. For the remaining interarrival time X we have EX = 1/10 hours, so 6 minutes after 12.00 hours. Hence at 12.06 hours. d. Let p = P(job has > 3 pages) = 1-e-2 (1+2+2+4/3) 0.143. Then {M(t)} is a Poisson process with rate 10p, so P(M(t) = m) = e10pt (10pt)m /m!. 2. a. Given the present, say Xn = i, the process will jump to 0 or i + 1 with probabi- lities that do not depend on Xn-1, Xn-2, . . .. Hence the process is a DTMC. It is irreducible (the path 0, 1, 2, 3, 4, 5, 6, 0 has positive probability so all states communicate), aperiodic (GCD(2,3,4,. . . )=1, so period of state 0, and hence all other states, is 1), and not transient but recurrent (finite closed class). Collections: Engineering