1 Quiz of ESE 520: Probability and Stochastic Processes
Problem 1. Caculate the probability: Pr(2 Z 4), where Z is a geometric distribution with p = 1/2, i.e., PZ (k) = 2k , k = 1, 2, . . .. Sol: Pr(2 Z 4) =Pr(Z = 2)+Pr(Z = 3)+Pr(Z = 4) = PZ (2) + PZ (3) + PZ (4) = 1/4 + 1/8 + 1/16 = 7/16. Problem 2. A communication channel has three states: A, B or C. The error probability for a bit transmitted through the channel is 0.05 when the channel is in state A, 0.02 when the channel is in state B, and 0.04 when the channel is in state C. Suppose that the channel is in state A, B, C with probablities 0.2, 0.3, 0.5 respectively. Compute the error probability when a bit is transmitted through this channel. Sol: Since the error is from A, B, or C, Pr(err) = Pr(err, A) + Pr(err, B) + Pr(err, C) = Pr(err|A)Pr(A) + Pr(err|B)Pr(B) + Pr(err|C)Pr(C) = 0.02 0.2 + 0.05 0.3 + 0.04 0.5 = 0.036 Problem 3. Suppose a pair of dice are rolled. (i) compute the probability that sum the of the two numbers is 4; Sol: Possible outcomes: (1, 3), (2, 2), (3, 1); Total outcomes: 6*6=36 Therefore, Pr(sum = 4) = 3/36 = 1/12. (ii) compute the probability that the sum of the two numbers is not less than 4; Sol: Use the formula: Pr(sum 4) = 1 Pr(sum 3) = 1 3/36 = 11/12 since the possible outcomes for the sum 3 are: (1, 1), (1, 2), (2, 1). (iii) compute the (conditional) probability that the sum of the two numbers is not less than 4 given that the number on the rst die is not greater than 2. (Hint: use the denition of conditional probability.) Sol: Let d1...