ECE 360_Hw 1 (2)

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University of Louisville *

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360

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Electrical Engineering

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Apr 3, 2024

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2

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ECE 360: Probabilistic Methods in Electrical and Computer Engineering Homework #1 1. [ 10 points ] Given P(A) = 0.5, P(B) = 0.46, and P(A B) = 2P(AB), what is P(AB)? 2P(AB)= .5 +.46-P(AB) 3P(AB) =.96 P(AB)=.32 2. [ 30 points ] Two fair dice are rolled. Let A be the event the sum is even and B be the event at least one of the numbers rolled is three. (a) [ 10 points ] What is the sample space? S ={(1,1),(1,2),(1,3) ...... (6,5),(6,6)} (b) [ 10 points ] Display the outcomes in a Karnaugh map in terms of events A and B. AB A Not A B AB B/A Not B A/B X B Not B (c) [ 10 points ] Determine P(AB). 18 outcumes with at least a single 3, 9 of those outcomes have an even sum. P(AB) = 9/36 =1/4 3. [ 10 points ] (Hajek’s book Problem 1.5) a) Not B B (3,2)(4,1)(4,3)(5,2) (6,1)(6,5) (1,4)(1,6)(2,3)(2,5) (3,4)(5,6) (1,2)(3,6) (4,5) (2,1)(5,4)(6,3) Not A (1,1)(2,2)(3,1)(4,4) (5,3)(5,5)(6,2)(6,4) (1,3)(2,6)(3,5)(4,6) (1,5)(2,4) (3,3)(4,2)(5,1) (6,6) A Not C C Not C b) P((A B)C) = (1,2)(3,6)(4,5)(1,3)(2,6)(3,5)(4,6)(1,5)(2,4) P((A B)C) = 9/36 =1/4 4: [ 50 points ] A certain town has three newspapers: A,B, and C . The probability distribution for which newspapers a person in the town reads is shown in the following Karnaugh map:
(a) [ 10 points ] Find the probability a person reads only newspaper A . =.01 (b) [ 10 points ] What is the probability a person reads at least two newspapers? =.03+.01+.07+.01=.12 (c) [ 10 points ] What is the probability a person doesn't read any newspaper? =.68 (d) [ 10 points ] If A and C are morning papers and B is an evening paper, what is the probability a person reads at least one morning paper plus an evening paper? =.07+.03+.01=.11 (e) [ 10 points ] What is the probability a person reads only one morning and one evening paper? =.07+.03=.10
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