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Unformatted text preview: EE 178 Handout #7 Probabilistic Systems Analysis Tuesday, February 5, 2008 Homework #3 Solutions 1. (10 points) a. We note that the probability of picking a person who tells the truth is 0 . 5, a person who lies is 0 . 3, and a person who does not give an answer is 0 . 2. Then, the probability of a single sequence, consisting of 10 people of each type, is (0 . 5) 10 (0 . 3) 10 (0 . 2) 10 . Since there are ( 30 10 10 10 ) sequences of length 30 and with 10 people of each type, the probability is given by parenleftbigg 30 10 10 10 parenrightbigg (0 . 5) 10 (0 . 3) 10 (0 . 2) 10 . b. We note that the probability of picking a person who lies is 0.3 and a person who does not lie is 0.7. Then, the probability of a single sequence, consisting of 12 liars and 18 others, is (0 . 3) 12 (0 . 7) 18 . Since there are ( 30 12 ) sequences of length 30 and with 12 liars and 18 others, the prob ability is given by parenleftbigg 30 12 parenrightbigg (0 . 3) 12 (0 . 7) 18 . 2. (20 points) a. The random variable W takes values in { , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 } . Its pmf is given as p W ( w ) = P { W = w } = P { : 7 summationdisplay i =0 i = w } = parenleftbigg 8 w parenrightbigg 1 2 8 , since there are ( 8 w ) 8dimensional binary vectors that have w 1s, for w { , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 } . Note that 8 summationdisplay w =0 p W ( w ) = 8 summationdisplay w =0 parenleftbigg 8 w parenrightbigg 1 2 8 = 1 2 8 8 summationdisplay w =0 parenleftbigg 8 w parenrightbigg = 1 2 8 (1 + 1) 8 = 1 , as it ought to be, since p W is a pmf. Going from the third to the fourth equation we have made use of the binomial fromula n summationdisplay k =0 parenleftbigg n k parenrightbigg a k b n k = ( a + b ) n b. It is X = braceleftbigg 1 , if there is even number of 1s , otherwise So p X (1) = P { X = 1 } = P {{ : 7 summationdisplay i =0 i = 2 k = even }} = P { : 7 summationdisplay i =0 i = 0 } + P( { : 7 summationdisplay i =0 i = 2 } + P { : 7 summationdisplay i =0 i = 4 } ) + P( { : 7 summationdisplay i =0 i = 6 } + P { : 7 summationdisplay i =0 i = 8 } = p W (0) + p W (2) + p W (4) + p W (6) + p W (8) = 1 2 8 parenleftbiggparenleftbigg 8 parenrightbigg + parenleftbigg 8 2 parenrightbigg + parenleftbigg 8 4 parenrightbigg + parenleftbigg 8 6 parenrightbigg + parenleftbigg 8 8 parenrightbiggparenrightbigg = 1 2 8 parenleftbigg 1 + 8 7 2 + 8 7 6 5 4 3 2 + 8 7 2 + 1 parenrightbigg = 1 2 8 (2 + 56 + 70) = 128 256 = 1 2 . Therefore, since p X (0) + p X (1) = 1, p X (0) = 1 2 , so p X ( x ) = braceleftbigg 1 / 2 , x = 1 1 / 2 , x = 0 c. It is Y = braceleftbigg 1 , if j = 1 , if j = 0 so p Y (1) = P { Y = 1 } = P { : j = 1 } ....
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This note was uploaded on 09/29/2008 for the course EE 178 taught by Professor Hewlett during the Spring '08 term at Stanford.
 Spring '08
 Hewlett

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