Solution - Assignment#1 - 2023

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University of California, Davis *

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114

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Civil Engineering

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Feb 20, 2024

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ECI-114 Probabilistic Systems Analysis for Civil Engineers, 2023 Department of Civil and Environmental Engineering University of California Davis Instructor: Prakash Singh Badal 3160 Ghausi Hall, psinghbadal@ucdavis.edu Homework due: July 3, 2023 Homework # 1: Statistics and Probability 1) 8 points, 2 each. Write down in set notation (intersection, union, conditional, etc.), the events represented by the White (or diagonal lines) shaded areas in the following Venn Diagrams. Example: represents the event A a) A Ո B b) [C U (A Ո B)]’ = C’ Ո (A ’ U B’) c) (B U C)

d) (A Ո B Ո C’) 2) 18 points, 3 each. A fair octahedral dice has 8 possible outcomes from 1 to 8. The sample space (S) is {1, 2, 3, 4, 5, 6, 7, 8}. Each outcome is equally likely. Let A be the event {3, 6}, and let B denote the event {2, 4, 6, 8}. Determine the following probabilities: a) P(A) = 2/8 b) P(B) = 4/8 c) P(A’) = 1 – P(A) = 6/8 d) P(A∩B) = 1/8 A∩B = {6} e) P(AUB) = 5/8 A U B = {2, 3, 4, 6, 8} f) P(A’∩B ) = 3/8 A’∩B = {2, 4, 8} 3) 18 points, 3 each. A sample space of random experiments is {r, s, t, u, v, w} with probabilities of 0.01, 0.45, 0.05, 0.04, 0.3, 0.15, respectively. Let A denote the event {r, s, u}, B denote the event {s, t, v}, and C denote the event {r, s, t}. Determine the following probabilities: a) P(A) = 0.01 + 0.45 + 0.04 = 0.50 b) P(B) = 0.45 + 0.05 + 0.30 = 0.80 c) P(C’) = 1 – P(C) = 1 – (0.01 + 0.45 + 0.05) = 1 – (0.51) = 0.49 d) P(A∩C) = 0.01 + 0.45 = 0.46 e) P(AUB) = 0.01 + 0.45 + 0.05 + 0.04 + 0.30 = 0.85 f) P(AUBUC) = 0.01 + 0.45 + 0.05 + 0.04 + 0.30 = 0.85 4) 12 points, 3 each . College of Engineering at UC Davis wants to place a large order for comfy chairs for different student spaces. A vendor was hired to fulfill the order. However, upon a preliminary inspection of the 1000 chairs given by the vendor, concerns about the quality of the chairs were raised. There were two main concerns: (1) the quality of the chairs (they need to meet their high-strength standard), and (2) the flaws within the chair remain below the detection/inspection limit. One of the student bodies was assigned for conducting the tests, and the results are summarized in the following table: Using the inspection results, help the company answer the following questions. a) For a randomly selected comfy chair, what is the probability that it meets the high strength standard? (688+216)/1000 = 904/1000 = 90.4% Meets High Strength Standard Yes No No Flaws Detected 688 75 Flaws Detected 216 21

b) For a randomly selected comfy chair, what is the probability that it meets the high strength standard and has no detectable flaws? 688/1000 = 68.8% c) For a randomly selected comfy chair, if it is known to have flaws detected, what is the probability that it has not met the high strength standard? 21/(216+21) = 21/237 = 8.86% d) If the event that a flaw is detected is denoted as event FD, and the event that a chair does not meet the high-strength standard is NHS, how would you write the probability you calculated in part (c) in set notation (intersection, union, conditional, etc.)? P(NHS∩FD)/P(FD) = P(NHS|FD) 5) 9 points, 3 each . Provide a reasonable description (e.g. S ={…}) of the sample space for the following random experiments. Define the events included for each experiment: a) Each of the 4 bridges on a highway corridor is analyzed and then classified as above or below specification. S = {YYYY, YYYN, YYNN, YYNY, YNNN, YNYY, YNYN, YNNY, NYYY, NYYN, NYNN, NNYY, NNYN, NNNY, NNNN, NYNY} b) In a simple game of darts, 3 darts are thrown at a board and can land in the center section or outer section of the board. S= {CCC, CCO, COO, OCC, COC, OOC, OCO, OOO} c) In a potato chip manufacturing plant, the quality control engineer draws a chip at random and classifies it as high quality or mediocre, and also as large, medium, or small in size. S= {Hl, Hm, Hs, Ml, Mm, Ms} 6) 8 points , 4 each . Suppose P(A|B) = 0.2, P(B) = 0.8, and P(A) = 0.64. a) What is P(A∩B)? = 0.2 * 0.8 = 0.16 b) What is P(B|A)? = 0.16/0.64 = 0.25 7) 15 points , 5 each. In the current Request for Proposals (RFP) process by the US Department of Transportation, civil engineering departments from 7 Universities are offering bids on 4 distinct/different research center funding contracts (which are different in the level of funding and the duration). Any University can only be awarded at most 1 grant to open a research center. a) Please explain how would you estimate the different ways in which the contracts could be awarded? Estimate the number. Estimate the number. Alt Ans 1: 7 * 6 * 5 * 4 = 840

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Alt Ans 2: b) Under the assumption that the simple events are equally likely (each University has the same probability of being awarded a contract), find the probability that University No. 2 is awarded one. Alt Ans 1: Using what we know about equally likely events: 4 contracts/7 universities = 4/7=0.571 Alt Ans 2: If we select all the ways to award contracts with a specific university in the selection, and divide by all the possible ways to assign fellowships: = 4/7 = 0.571 c) Suppose that Universities 5 and 6 have submitted bids, if the contracts are awarded at random by the DOT, find the probability that both of these Universities receive contracts. (Hint, you can do this by analyzing the conditional probabilities, or by using counting methods) Alt Ans 1: Simplest approach: University 5 (U5) has the probability of winning a contract: P(U5) = (number of contracts)/(number of universities)=4/7 Now that U5 has already won a contract, university 6 (U6) has some probability of winning a contract: P(U6) = (number of remaining contracts)/(number of remaining universities) =3/6 So the probability of both winning a contract is the probability of both universities getting a contract with the probability given above P(both students getting a contract) = P(U5)P(U6) = (4/7)(3/6) = 0.2857 Alt Ans 2: Thinking about it in terms of conditional probability: We want to know the probability of both universities 5 and 6 getting a contract, or P(U 5∩ U6). We also know: P(U 5∩ U6) = P(U 6∩ U5) = P(U6|U5)(U 5). From above in our “simple approach” we calculated the probability of U6 getting a contract given that U5 got a contract, (number of remaining contracts)/(number of remaining Us), this is then a conditional probability. So, P(U 6∩ U5) = P(U6|U5)(U5) = [(number of contracts - 1)/(number of Us – 1)] * [(number of contracts)/(number of Us)]= (3/6)(4/7) = 0.2857 Alt Ans 3: Using our understanding of Permutations We could also figure out this problem based on the relative frequency of all the ways to distribute contracts that include Us 5 and 6, and the total number of ways to distribute contracts: The total number of ways to choose 2 contracts out of 4 (one for U5 and one for U6): U5 U6 __ __, U6 U5 __ __, __ U5 U6 __, __ U6 U5 __, U5 __ U6 __, …, etc. So in terms of writing this as a permutation: 4 2 P But there still remain 2 other contracts to give out, and 5 Us left to assign them to, and for example, U5 U6 U1 U2 is NOT the same as U5 U6 U2 U1, we also need to calculate all the ways the remaining two contracts could be assigned, or 5 2 P .

So, we have the total number of ways to assign U5 and U6 contracts: 4 2 P x 5 2 P . To solve the problem, all you have to do is calculate the proportion (or relative frequency) of the ways Us 5 and 6 can both be assigned contracts, divided by all the ways to assign contracts: ( 4 2 P x 5 2 P )/ 7 4 P = 240/840 = 0.2857 8) 12 points. Prove the following DeMorgan’s law graphically: (A  B)’ = A’  B’

Solution - Assignment#1 - 2023

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