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Unformatted text preview: Class Test 1 ECE502 — Analysis of Probabilistic Signals and Systems (Fall .2008)
Instructor: Professor A. M. Wyglinski September 22, 2008
08:00pm — 08:50pm Instructions The duration of this class test is 50 minutes. This clms test consists of three (3) questions, each of which has a‘ diﬂ'erent point value.
This class test is closed book. One letter size sheet with handwritten notes on both sides is permitted. Calculators are permitted. Please write neatly and legibiy. Each point in this class test corresponds to one minute of doing the test. Therefore, this
test is out of 50 points. Budget your time accordingly. Attempt all questions; clearly Show all your work and not just the final answer.
Collaborating with other students during the class test is strictly prohibited.
Hand in all material to the instructor at the end of 50 minutes. Good luck! ——.———— '— Last Name First Name Student ID Mailbox Question 1: Axioms of Probability [10 Points] A dart board has an area of 143 square inches. with a “bull‘s eye" at the center of area 1 square inch.
The rest of the board is divided into 20 sectors numbered 1, 2, 3. ..., 20. There is also a triple ring
with an area of 10 square inches and a double ring with an area of 15 square inches. Note that these
area values are all rounded to the nearest integer. S Figure 1 for the schematic of a. cenventional dart board. Ndi‘c. d'luQL c‘rL avgLad ggr'fb— (enciao] b..”‘.r (ye Lbf‘ Suppose you throw a dart at random at the board. What is the probability the. you will get: ' "Ch"J3. (
(a) 2 points: Double 14'? {:3 g
(b) 3 points: 14 but not a double? ’f‘f‘u/IC ,.
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(c) 2 points: Triple or the bull’s eye? ‘ {755
(d) 3 points: An even number or a double? 2° HINT: Use the following symbolic notation when solving this problem: F = {14}. D = {double}1
T = {triple}. B : {bull’s eye}, and E = {even}. Bull's Eye Double Ring Figure 1: Schematic of a conventional dart board. a M. a @ Pom F) : ”55””C 2700..
PCH D) = Pm — Porno)
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._ 97/) Z5 51,; Question 1 (Continued) Question 2: Combinatorics [25 Points] Suppose we have an urn containing 10 white balls, 10 red balls, and 10 black balls. You draw 5 balls at random without replacement. Suppose now that we deﬁne the following events: R = {no red balls}
W = {no white balls}, and B = {no blue balls}. Assume that ‘7 (I Incl/5 9‘4 equcil‘, (‘4‘;{7 . (a) 2 points: What is the binomial coefﬁcient expression (2) equal to in terms of factorials? (b) 5 points: What is the probability that when choosing 5 balls from the urn that none of them
are red, i.e., PW)? HINT: Use the binomial coefﬁcient expression in part (a). (c) 6 points: What is the probability that when choosing 5 balls from the urn that none of them
are red or white, i.e., P(R I”) W)? (d) 5 points: Expand the general expression for PUB U W U B) when the events R, W, and B are
not necessarily disjoint. Express result in terms of the probabilities of the individual events and
the intersection of two events. 6. «a; c Stun—1, POQ/l W“ 3) =' O . 9 uh, I. : P(£n wn :3) =0?
(e) 7 pointjﬂssuming that P(R) = P(W) : P(‘B), what is the probability that you do not get all
colors? HINT: See part ((1) /
C9 02 t a, .l I'VHJ ,ch Question 2 (Continued) @ 19(KU WUB): 3(a) —3P(KmM/)
:2 3((2‘0— (ﬁn/(£7 40,325 Question 2 (Continued) Question 3: Conditional Probability [15 Points] Suppose we have a. ternary communication system that transmits the symbols :50 = 0, 3:1 : 1, and
;::2 = 2 across an error—prone channel (see Figure 2}. These three input symbols to the channel are
all equally likely to occur. Suppose the probability that an input symbol to the channel did not suffer
rm error is 1 — E. Furthermore, suppose the only error conditions possible in the channel are when an
input symbol of mu : 0 becomes an output symbol of y1 = 1, an input symbol of :51 = 1 becomes an
output symbol of 1J2 = 2, and an input symbol of m2 = 2 becomes an output symbol of yo = 0. The
probability that such an error W111 occur 13 E. who... a L“we c a /‘~3
(a) 1 points: What is the expression for the Law of Total Probability? e M A— Tlu: i’ p ._+.+,_‘ b an L 1c 9
. _ . H 7 , . 7 e C e: o ”Hfu/
(b) 1 paints. What is the expression for Bayes Rule. end441...; e 'b 5'" I}, 9
(c) 3 points: Find the conditional probabilities P(ya$u'), P(y0:r1), P_(yg_:cg), P('y1$g), P(y1;1:1), ( ("K
P(y12'2), P(:ijgfﬂg), P(y2lﬂ:1), and P(y23:2). HINT. 110015. at. Figure 2 ”‘6 ﬂ ’3 r
(d) 6 points: Find the probabilities of the outputs P(y0), P(y1), and P(y2). HINT: Use thesolutions ”as: , 6K
from part (a). 1‘ .9
(e) 4 points: Suppose y1 = l is observed at the output. What is the probability that the input was 5.34
a m0 = 0‘? °¢..J Figure 2: Ternary communication system channel. 6) We) :FGA—WQHEJ .1. ..+ WMQij
Poe/15;) PKA/QWKB
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 Spring '09
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