ece226_hw1_part2

ece226_hw1_part2 - P [ C ∪ D c ] = P [ C ] + P [ D ] −...

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Probability and Stochastic Processes Homework 1 Solutions 29th Jan, Spring’10 Problem Solutions : Yates and Goodman, 1.6.4 and 1.6.5 Problem 1.6.4 Solution (a) Since A B = , P [ A B ] = 0. To fnd P [ B ], we can write P [ A B ] = P [ A ] + P [ B ] P [ A B ] (1) 5 / 8 = 3 / 8 + P [ B ] 0 . (2) Thus, P [ B ] = 1 / 4. Since A is a subset oF B c , P [ A B c ] = P [ A ] = 3 / 8. ±urthermore, since A is a subset oF B c , P [ A B c ] = P [ B c ] = 3 / 4. (b) The events A and B are dependent because P [ AB ] = 0 n = 3 / 32 = P [ A ] P [ B ] . (3) (c) Since C and D are independent P [ CD ] = P [ C ] P [ D ]. So P [ D ] = P [ CD ] P [ C ] = 1 / 3 1 / 2 = 2 / 3 . (4) In addition, P [ C D c ] = P [ C ] P [ C D ] = 1 / 2 1 / 3 = 1 / 6. To fnd P [ C c D c ], we frst observe that P [ C D ] = P [ C ] + P [ D ] P [ C D ] = 1 / 2 + 2 / 3 1 / 3 = 5 / 6 . (5) By De Morgan’s Law, C c D c = ( C D ) c . This implies P [ C c D c ] = P [( C D ) c ] = 1 P [ C D ] = 1 / 6 . (6) Note that a second way to fnd P [ C c D c ] is to use the Fact that iF C and D are independent, then C c and D c are independent. Thus P [ C c D c ] = P [ C c ] P [ D c ] = (1 P [ C ])(1 P [ D ]) = 1 / 6 . (7) ±inally, since C and D are independent events, P [ C | D ] = P [ C ] = 1 / 2. (d) Note that we Found P [ C D ] = 5 / 6. We can also use the earlier results to show
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Unformatted text preview: P [ C ∪ D c ] = P [ C ] + P [ D ] − P [ C ∩ D c ] = 1 / 2 + (1 − 2 / 3) − 1 / 6 = 2 / 3 . (8) (e) By Defnition 1.7, events C and D c are independent because P [ C ∩ D c ] = 1 / 6 = (1 / 2)(1 / 3) = P [ C ] P [ D c ] . (9) 1 Problem 1.6.5 Solution For a sample space S = { 1 , 2 , 3 , 4 } with equiprobable outcomes, consider the events A 1 = { 1 , 2 } A 2 = { 2 , 3 } A 3 = { 3 , 1 } . (1) Each event A i has probability 1 / 2. Moreover, each pair of events is independent since P [ A 1 A 2 ] = P [ A 2 A 3 ] = P [ A 3 A 1 ] = 1 / 4 . (2) However, the three events A 1 ,A 2 ,A 3 are not independent since P [ A 1 A 2 A 3 ] = 0 n = P [ A 1 ] P [ A 2 ] P [ A 3 ] . (3) 2...
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This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.

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ece226_hw1_part2 - P [ C ∪ D c ] = P [ C ] + P [ D ] −...

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