hw2-solution

hw2-solution - Probability and Stochastic Processes A...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 1.5.5 1.6.4 1.7.5 1.8.1 1.8.3 1.8.4 1.8.6 1.9.2 1.9.4 1.10.1 and 1.10.2 Problem 1.5.5 Solution The sample outcomes can be written ijk where the first card drawn is i , the second is j and the third is k . The sample space is S = { 234 , 243 , 324 , 342 , 423 , 432 } . (1) and each of the six outcomes has probability 1 / 6. The events E 1 , E 2 , E 3 , O 1 , O 2 , O 3 are E 1 = { 234 , 243 , 423 , 432 } , O 1 = { 324 , 342 } , (2) E 2 = { 243 , 324 , 342 , 423 } , O 2 = { 234 , 432 } , (3) E 3 = { 234 , 324 , 342 , 432 } , O 3 = { 243 , 423 } . (4) (a) The conditional probability the second card is even given that the first card is even is P [ E 2 | E 1 ] = P [ E 2 E 1 ] P [ E 1 ] = P [243 , 423] P [234 , 243 , 423 , 432] = 2 / 6 4 / 6 = 1 / 2 . (5) (b) The conditional probability the first card is even given that the second card is even is P [ E 1 | E 2 ] = P [ E 1 E 2 ] P [ E 2 ] = P [243 , 423] P [243 , 324 , 342 , 423] = 2 / 6 4 / 6 = 1 / 2 . (6) (c) The probability the first two cards are even given the third card is even is P [ E 1 E 2 | E 3 ] = P [ E 1 E 2 E 3 ] P [ E 3 ] = 0 . (7) (d) The conditional probabilities the second card is even given that the first card is odd is P [ E 2 | O 1 ] = P [ O 1 E 2 ] P [ O 1 ] = P [ O 1 ] P [ O 1 ] = 1 . (8) (e) The conditional probability the second card is odd given that the first card is odd is P [ O 2 | O 1 ] = P [ O 1 O 2 ] P [ O 1 ] = 0 . (9) 1 Problem 1.6.4 Solution (a) Since A ∩ B = ∅ , P [ A ∩ B ] = 0. To find P [ B ], we can write P [ A ∪ B ] = P [ A ] + P [ B ]- P [ A ∩ B ] (1) 5 / 8 = 3 / 8 + P [ B ]- . (2) Thus, P [ B ] = 1 / 4. Since A is a subset of B c , P [ A ∩ B c ] = P [ A ] = 3 / 8. Furthermore, 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 6 = 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 find P [ C c ∩ D c ], we first 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 find 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) Finally, since C and D are independent events, P [ C | D ] = P [ C ] = 1 / 2....
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This note was uploaded on 08/25/2008 for the course EEE 350 taught by Professor Duman during the Spring '08 term at ASU.

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hw2-solution - Probability and Stochastic Processes A...

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