ECE109SU2017_Final_Solutions.pdf

# ECE109SU2017_Final_Solutions.pdf - ECE 109 Summer Session...

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ECE 109 - Summer Session II 2017 Final Exam University of California, San Diego September 8th, 2017 8:00am to 10:59am Center Hall 119 Name: SOLUTIONS Student ID Number: Instructions: No calculators or other electronic devices allowed. 1 page (both sides allowed) of notes allowed, but should be handed in with the exam. Write your answers in the spaces provided. 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. By signing below, you acknowledge that you have read the instructions and you have neither given nor received unauthorized aid on this exam: Signature: Scores P1 / 8 P2 / 6 P3 / 6 P4 / 7 P5 / 8 P6 / 9 P7 / 14 P8 / 8 P9 / 6 P10 / 8 P11 / 8 P12 / 12 / 100 1

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Problem 1: [8 points] In how many ways can 8 people be seated in a row if (a) (1 point) there are no restrictions on the seating arrangement? Solution: There are 8! ways of permuting 8 people. (b) (1 points) persons A and B must sit next to each other? Solution: There are 2 ways for person A to be seated in an end seat. If person B is seated next to per- son A, there are 6! ways of permuting the remaining 6 people. There are 6 ways for person A to be seated in a middle seat, and for each of these 6 ways of seating person A, there are 2 ways of seating person B next to person A. There are then 6! ways of permuting the remaining 6 people. | A | = | A B | + | A B c | = (2)(6!) + (2)(6)(6!) = (14)(6!) (c) (2 points) there are 4 men and 4 women, and no two men or two women can sit next to each other? Solution: If the men and women are not seated by each other, the arrangement must be either (i) M W M W M W M W (ii) W M W M W M W M There are (4!)(4!) ways of permuting the 4 men and the 4 women in each of these arrangements, so there are 2(4!)(4!) ways of having no two men or no two women seated next to one another. 2
(d) (2 points) there are 5 men and 3 women, and all men must sit together? Solution: There are 4 ways of arranging the group of 5 men. M M M M M W W W W M M M M M W W W W M M M M M W W W W M M M M M For each of these 4 arrangements, there are 3! ways of permuting the women and there are 5! ways of permuting the men. Thus there are (4)(5!)(3!) ways of seating all 5 men together. (e) (2 points) there are 4 married couples and each couple must sit together? Solution: When the 4 couples must sit together, there are 4! ways of permuting the couples, and each person in the couple could be in one of 2 seats, so there are 2 4 4! such arrangements. 3

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Problem 2: [6 points] When answering multiple-choice questions, a student knows the answer with probability 3 4 and guesses with probability 1 4 . Assume that a student who guesses at the answer will be correct with probability 1 5 . What is the probability that the student knew the answer given that he or she answered a question correctly?
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• Spring '08
• KennethZeger
• Probability theory, probability density function, Probability mass function, Xi Yi

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