{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hw3sol

Hw3sol - EE 178 Handout#7 Probabilistic Systems Analysis...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 { ω : 7 summationdisplay i =0 ω i = w } = parenleftbigg 8 w parenrightbigg · 1 2 8 , since there are ( 8 w ) 8-dimensional binary vectors that have w 1’s, 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 1’s , otherwise So p X (1) = P { X = 1 } = P {{ ω : 7 summationdisplay i =0 ω i = 2 k = “even”’ }} = P { ω : 7 summationdisplay i =0 ω i = 0 } + P( { ω : 7 summationdisplay i =0 ω i = 2 } + P { ω : 7 summationdisplay i =0 ω i = 4 } ) + P( { ω : 7 summationdisplay i =0 ω i = 6 } + P { ω : 7 summationdisplay i =0 ω 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 ω j = 1 , if ω j = 0 so p Y (1) = P { Y = 1 } = P { ω : ω j = 1 } ....
View Full Document

{[ snackBarMessage ]}

Page1 / 8

Hw3sol - EE 178 Handout#7 Probabilistic Systems Analysis...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online