Midterm05

Midterm05 - University of California, San Diego Department...

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Unformatted text preview: University of California, San Diego Department of Electrical and Computer Engineering ECE109 - Spring 2005 Midterm Exam - SOLUTIONS Wednesday, May 4, 2005 5:00pm- 6:20pm Location: HSS 2250 K. Zeger Name Your UCSD ID Number Signature INSTRUCTIONS This exam is open book and open notes. No calculators, laptop computers, or other electronic devices are allowed. Write your answers in the spaces provided. Show all your work. If you need extra space, please use the back of the previous page. Partial credit will be given only for substantial progress on a problem. Zero credit will be given for correct answers that lack adequate explanation of how they were obtained. There is a maximum total of 30 points on this exam. Simplify your answers as much as possible and leave answers as fractions, not was decimal numbers. GRADING 1. 10 points 2. 10 points 3. 10 points TOTAL (30 points) ... 1 Problem 1 (10 points) Al, Bo, and Cy tell the truth according to certain probabilities. The probability that either Al or Cy lies is and the probability both Al and Cy lies is . The probability that Cy lies but Al tells the truth is , and the probability Al and Bo lie while Cy tells the truth is . The probability that all three of them lie is . What is the probability that all but Al are truthful, given that Al and Cy are not both truthful? ¤ , denote the events that Al, Bo, and Cy are lying, respectively. Then we are §¦¥ &'¤ #%$¥  §¦ ¢ #§  ¥ ¨  © ¡ #$¥ © ¨  § ©   § ¨ #"!¥ £ ¥ ¨¨   § ¦ © © is a disjoint union, we have § ¥ § ¥  § ¦ ¥  § ¦ ¥  § )42$"3#%)3#210)(¥ ¢ Therefore, since ¤  § ¦ © 1X#¥ ¨ § U)!¥ § U)!¥ & E¢ C ¡ C @ W(UV!B£16    §  $¥¨ ¨  TP!S¥RE  § ¨ $B¥@ ¨ 7#P!¥ R §  ¥ ¨ ¦ ©  E § @ © ¦ ©   § Q ¦ © © H § P!¥ I §  $¥ ¦ &E¢ C ¡ C @ GF!#!DB£A6 5#§  ¥ ¨ 9#$¥ ¨ 8  %¥ ¨ 7#"!¥ ¨ 5  §  $¥ ¨  © 6 § © 6 § ¦ © 6 § ©  ¦© So, the quantity we are asked to find is (using the fact that Note that £ ¡ SOLUTION: Let , given: ) was extraneous information that was not needed to solve the problem. 2 & d e c C 6E@ U W A6 E C H 2 6  ¦ C @ %QI   C¤ E@ 5 ¤I ¤ ¢ QP ¥ ©¡  8 754 6 3 @ E @ E a F 6  ¨¦ c b §8 7 P £© E @ Y  FR§8 [email protected]  &  R§8 754  U £© ¨¦ E 3 FE 6 3 C C Y£X  FR§8 67534 WV £© 0 U ¨¦ E C H 7 Y  R§8 [email protected]  X WV `© 0U ¨¦ FE 6 3 C  G§8 [email protected] FE 6 C ¤ TI T  ¨¦ £© H  RB§8 [email protected] FE 6 93 or ! 3 B  R8 754 FE 6 3 C ¤ IE @ ¢ QP5¥¤©¡ 6 E £©@  ¨¦ C SB H 7  G§8 754 FE 6  3 C ' 0!¦ 1)( iff  8 [email protected] 6 93   C DB 6  6 A (!  ¦  E W @  5¥¤£¡ [email protected] © ¨ & %!  ¦ iff (by the ¤ SOLUTION: Note that quadratic formula). Thus,   ¨¦   £©§ 6 $ #" ¦! © ¨ Let be a random variable whose probability density function is Find . for all . Problem 2 (10 points) Problem 3 (10 points) satisfying all of the following three properties: Find a random variable © ¢ ¤ © ¨ £X8S2 ¨ ¦ 5 2 2 0 ¡ © ¨ ¤ © 58D ¨ 0 . . [email protected] 5¥¤#¨  7¤ 2 You must precisely specify either the probability mass function bility density function if is continuous. if [email protected] ¡ [email protected] for all E W£¨ (iii) (ii) for all real numbers . ¤ (i) is discrete, or the proba- [email protected] 5¥¤¡ SOLUTION: Property (i) tells us the random variable is continuous, not discrete. Property (ii) tells us the positive portion of the pdf of is confined to the range . Property (iii) says that the area under the pdf from to should be times the area under the same curve from to . This can be accomplished by having the pdf increase linearly with (i.e. triangle shaped), as: for and otherwise. This is because (area of triangle), so §  8(2 ¤ © ¤ ¨ ¤  0 § ##£¨  E3¤ § X8D2 @ ¤  5¥¤©¡ [email protected] 0 1 & 8'2 ¨  ¤ © ¢ ¤ § © ¨  ¡ ©0 © ¦ [email protected] 5¤ #¡ § W£Q@  EFX6  @ E  ¨ @  ¤ I  E¨ 0  ¡ RW`©¦§ T¤ B § $¦  4  £¨  ¤ ¢ and obtain 0 ¤ ¤ ¤ . ¤ §  5¥¤©¡ [email protected] ¨  £#0§ so  ¡ ©0 © ¦ We solve for the constant ...
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This test prep was uploaded on 04/07/2008 for the course ECE 109 taught by Professor Kennethzeger during the Spring '08 term at UCSD.

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