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Probset3_sol - Probability and Stochastic Processes...

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Probability and Stochastic Processes Homework 3 Solutions 12th Feb, Spring 2010 Problem Solutions : Yates and Goodman, 2.2.1 2.2.3 2.2.6 2.2.8 2.3.1 2.3.3 2.3.4 2.3.7 2.3.10 2.3.11 2.3.13 2.4.1 2.4.3 and 2.4.5 Problem 2.2.1 Solution (a) We wish to find the value of c that makes the PMF sum up to one. P N ( n ) = braceleftbigg c (1 / 2) n n = 0 , 1 , 2 0 otherwise (1) Therefore, 2 n =0 P N ( n ) = c + c/ 2 + c/ 4 = 1, implying c = 4 / 7. (b) The probability that N 1 is P [ N 1] = P [ N = 0] + P [ N = 1] = 4 / 7 + 2 / 7 = 6 / 7 (2) Problem 2.2.3 Solution (a) We must choose c to make the PMF of V sum to one. 4 summationdisplay v =1 P V ( v ) = c (1 2 + 2 2 + 3 2 + 4 2 ) = 30 c = 1 (1) Hence c = 1 / 30. (b) Let U = { u 2 | u = 1 , 2 ,... } so that P [ V U ] = P V (1) + P V (4) = 1 30 + 4 2 30 = 17 30 (2) (c) The probability that V is even is P [ V is even] = P V (2) + P V (4) = 2 2 30 + 4 2 30 = 2 3 (3) (d) The probability that V > 2 is P [ V > 2] = P V (3) + P V (4) = 3 2 30 + 4 2 30 = 5 6 (4) 1
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Problem 2.2.6 Solution The probability that a caller fails to get through in three tries is (1 p ) 3 . To be sure that at least 95% of all callers get through, we need (1 p ) 3 0 . 05. This implies p = 0 . 6316. Problem 2.2.8 Solution From the problem statement, a single is twice as likely as a double, which is twice as likely as a triple, which is twice as likely as a home-run. If p is the probability of a home run, then P B (4) = p P B (3) = 2 p P B (2) = 4 p P B (1) = 8 p (1) Since a hit of any kind occurs with probability of .300, p + 2 p + 4 p + 8 p = 0 . 300 which implies p = 0 . 02. Hence, the PMF of B is P B ( b ) = 0 . 70 b = 0 0 . 16 b = 1 0 . 08 b = 2 0 . 04 b = 3 0 . 02 b = 4 0 otherwise (2) Problem 2.3.1 Solution (a) If it is indeed true that Y , the number of yellow M&M’s in a package, is uniformly distributed between 5 and 15, then the PMF of Y , is P Y ( y ) = braceleftbigg 1 / 11 y = 5 , 6 , 7 ,..., 15 0 otherwise (1) (b) P [ Y < 10] = P Y (5) + P Y (6) + · · · + P Y (9) = 5 / 11 (2) (c) P [ Y > 12] = P Y (13) + P Y (14) + P Y (15) = 3 / 11 (3) (d) P [8 Y 12] = P Y (8) + P Y (9) + · · · + P Y (12) = 5 / 11 (4) Problem 2.3.3 Solution Whether a hook catches a fish is an independent trial with success probability h . The the number of fish hooked, K , has the binomial PMF P K ( k ) = braceleftbigg ( m k ) h k (1 h ) m k k = 0 , 1 ,...,m 0
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