# soln2 - Probability and Stochastic Processes A Friendly...

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Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,1.5.1 1.5.3 1.5.4 1.5.6 1.6.6 1.7.1 1.7.3 1.7.6 1.7.9 1.8.2 1.8.4 and 1.8.5 Problem 1.5.1 probability. (a) Note that the probability a call is brief is P [ B ] = P [ H 0 B ]+ P [ H 1 B ]+ P [ H 2 B ] = 0 . 6 . The probability a brief call will have no handoffs is P [ H 0 | B ] = P [ H 0 B ] P [ B ] = 0 . 4 0 . 6 = 2 3 (b) The probability of one handoff is P [ H 1 ] = P [ H 1 B ]+ P [ H 1 L ] = 0 . 2. The probability that a call with one handoff will be long is P [ L | H 1 ] = P [ H 1 L ] P [ H 1 ] = 0 . 1 0 . 2 = 1 2 (c) The probability a call is long is P [ L ] = 1 - P [ B ] = 0 . 4. The probability that a long call will have one or more handoffs is P [ H 1 H 2 | L ] = P [ H 1 L H 2 L ] P [ L ] = P [ H 1 L ]+ P [ H 2 L ] P [ L ] = 0 . 1 + 0 . 2 0 . 4 = 3 4 Problem 1.5.3 Since the 2 of clubs is an even numbered card, C 2 E so that P [ C 2 E ] = P [ C 2 ] = 1 / 3. Since P [ E ] = 2 / 3, P [ C 2 | E ] = P [ C 2 E ] P [ E ] = 1 / 3 2 / 3 = 1 / 2 The probability that an even numbered card is picked given that the 2 is picked is P [ E | C 2 ] = P [ C 2 E ] P [ C 2 ] = 1 / 3 1 / 3 = 1 Problem 1.5.4 Define D as the event that a pea plant has two dominant y genes. To find the conditional proba- bility of D given the event Y , corresponding to a plant having yellow seeds, we look to evaluate P [ D | Y ] = P [ DY ] P [ Y ] 1

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Note that P [ DY ] is just the probability of the genotype yy . From Problem 1.4.3, we found that with respect to the color of the peas, the genotypes yy , yg , gy , and gg were all equally likely. This implies P [ DY ] = P [ yy ] = 1 / 4 P [ Y ] = P [ yy , gy , yg ] = 3 / 4 Thus, the conditional probability can be expressed as P [ D | Y ] = P [ DY ] P [ Y ] = 1 / 4 3 / 4 = 1 / 3 Problem 1.5.6 The problem statement yields the obvious facts that P [ L ] = 0 . 16 and P [ H ] = 0 . 10. The words “10% of the ticks that had either Lyme disease or HGE carried both diseases” can be written as P [ LH | L H ] = 0 . 10 (a) Since LH L H , P [ LH | L
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