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hw2-solution

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

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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
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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 ] - 0 . (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. (d) Note that we found P [ C D ] = 5 / 6. We can also use the earlier results to show P [ C D c ] = P [ C ] + P [ D ] - P [ C D c ] = 1 / 2 + (1 - 2 / 3) - 1 / 6 = 2 / 3 . (8) (e) By Definition 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) 2
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Problem 1.7.5 Solution The P [ - | H ] is the probability that a person who has HIV tests negative for the disease. This is referred to as a false-negative result. The case where a person who does not have HIV but tests positive for the disease, is called a false-positive result and has probability P [+ | H c ]. Since the test is correct 99% of the time, P [ -| H ] = P [+ | H c ] = 0 . 01 . (1) Now the probability that a person who has tested positive for HIV actually has the disease is P [ H | +] = P [+ , H ] P [+] = P [+ , H ] P [+ , H ] + P [+ , H c ] . (2) We can use Bayes’ formula to evaluate these joint probabilities.
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