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Material Type: Exam; Professor: Ansari; Class: Probability and Random Processes for Engineers; Subject: Electrical and Computer Engr; University: University of Illinois - Chicago; Term: Spring 2014;
Typology: Exams
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ECE 341 Probability and Random Processes for Engineers Spring 2011
FINAL EXAM grade breakdown
Let X and Y be discrete random variables with joint probability mass function given by
PXY (x, y) =
c(1 + x + y) {x = 0, 1 , 2 } ∩ {y = 0, 1 , 2 } ∩ {x + y ≤ 2 } 0 otherwise
(a) 1.5 points. c = 1/14. (b) 1.5 points. P [X < Y ] = 5/14. (c) 2 points. E[|X − 1 |(3 − Y )^2 ] = 47/14. (d) 3 points. W = X + Y. Find the probability mass function PW (.) of the random variable W.
PW (w) =
1 / 14 w = 0 4 / 14 w = 1 9 / 14 w = 2 0 otherwise .
Let X denote the arrival time of the first telephone call at a switch, and Y denote the arrival time of the second telephone call. Note that Y ≥ X. Let X and Y have joint probability density function
fXY (x, y) =
3 x { 0 ≤ x ≤ y} ∩ {y ≤ 2 − x} 0 otherwise
(a) 1 point. Sketch of region in the (x, y) plane where fXY (x, y) is non-zero.
(^00) 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
1
2
(b) 2 points. E[XY ] = 0.5. (c) 4 points Event B = “X + Y ≤ a”,0 < a < 2. P [B] = a^3 / 8 .
a) Find E[T ] = 36E[Xk] = 144 and V ar[T ] = σ T^2 = 36V ar[Xk] = 144. Therefore σT = 12. b) Let event A =“126 < T ≤ 156” and event B =“T > 132”. Now A ∩ B = “132 < T ≤ 156”, So P [A ∩ B] = P [132 < T ≤ 156] = Φ( 156 − 12144 ) − Φ( 13212 −^144 ) = Φ(1) − Φ(−1). Also B = “132 < T ≤ 156”, so P [B] = 1 − P [T ≤ 132] = 1 − Φ( 13212 −^144 ) = 1 − Φ(−1).
The answer should be in the form of a ratio involving Φ(.) values..
