By Roy D. Yates, David Goodman
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Extra info for Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers (solution manual)
001 is the probability a packet is corrupted. 999 (2) (b) Let Y denote the number of packets received in error out of 100 packets transmitted. 999)100−y y = 0, 1, . . 1 (4) (c) Let L equal the number of packets that must be received to decode 5 packets in error. L has the Pascal PMF l−1 5 l−5 l = 5, 6, . . 001 (6) (d) If packet arrivals obey a Poisson model with an average arrival rate of 1000 packets per second, then the number N of packets that arrive in 5 seconds has the Poisson PMF PN (n) = 5000n e−5000 /n!
2 (1) (2) In this case, the probability the system works is P [WI ] = P [W1 W2 W3 ∪ W4 ] P [W5 ∪ W6 ] = 1 − q2 (5 − 4q + q 2 ) (1 − q 2 ). 2 (3) II Replace component 4 In this case, P [W1 W2 W3 ] = (1 − q)3 , q P [W4 ] = 1 − , 2 P [W5 ∪ W6 ] = 1 − q 2 . (4) q q + (1 − q)3 . 2 2 (5) This implies P [W1 W2 W3 ∪ W4 ] = 1 − (1 − P [W1 W2 W3 ])(1 − P [W4 ]) = 1 − In this case, the probability the system works is P [WII ] = P [W1 W2 W3 ∪ W4 ] P [W5 ∪ W6 ] = 1 − 30 q q + (1 − q)3 (1 − q 2 ). 2 2 (6) III Replace component 5 In this case, P [W1 W2 W3 ] = (1 − q)3 , P [W4 ] = 1 − q, P [W5 ∪ W6 ] = 1 − q2 .
0 otherwise. PK (k) = (1) For any integer k ≥ 1, the CDF obeys k FK (k) = k pq j−1 = 1 − q k . PK (j) = j=1 (2) j=1 Since K is integer valued, FK (k) = FK ( k ) for all integer and non-integer values of k. ) Thus, the complete expression for the CDF of K is 0 k < 1, (3) FK (k) = k 1 − (1 − p) k ≥ 1. 5 Solution Since mushrooms occur with probability 2/3, the number of pizzas sold before the ﬁrst mushroom pizza is N = n < 100 if the ﬁrst n pizzas do not have mushrooms followed by mushrooms on pizza n + 1.