EE114 Exams

3/15/2023 10 Q. Calculate MGF for the given RV: 10 Points Which one of the following pdf’s are independent? Why? Independent f(x,y) = 4 𝑥 ଶ 𝑦 ଷ f(x) = a 𝑥 ଶ & f(y) = b 𝑦 ଷ , where ab = 4 Not independent f(x,y) = ½ ( 𝑥 ଷ 𝑦 + 𝑥 𝑦 ଷ ) because f(x,y) can not be factored into a product of x & y Independent f(x,y) = 6 𝑒 ିଷ௫ିଶ f(x) = a 𝑒 ିଷ௫ & f(y) = b 𝑒 ିଶ௬ , ab=6 98 101

3/15/2023 14 Q. Given the following PMFs, find E[X], Var[X], E[X+Y], Var[X+Y] and .. . Q. Given independent random variables X & Z, and Y=X+Z. E[X] = E[Z] = 0 Var[X] = 1, Var[Z] = 16 Find the correlation coefficient of X and Y. Are X & Y independent? 118 121

3/15/2023 17 Q. Suppose that a fair die is rolled just once. Let Y be 1 if 3 comes up and zero otherwise. Let Z be 1 if 2 comes up and zero otherwise. Compute the covariance Cov(Y, Z) and the variances Var(Y ) and Var(Z). E[Z] =?, E[Y] = ?, E[YZ] = ? Y=1 prob 1/6 Y=0 prob 5/6 E[Y] = 1 x 1/6 + 0 x 5/6 = 1/6 E[Z] = 1/6 E[YZ] = ? Q. Suppose that a fair die is rolled just once. Let Y be 1 if 3 comes up and zero otherwise. Let Z be 1 if 2 comes up and zero otherwise. Compute the covariance Cov(Y, Z) and the variances Var(Y ) and Var(Z). E[Z] =?, E[Y] = ?, E[YZ] = ? Y=1 prob 1/6 Y=0 prob 5/6 E[Y] = 1 x 1/6 + 0 x 5/6 = 1/6 E[Z] = 1/6 1 2 3 4 5 6 Z 0 1 0 0 0 0 Y 0 0 1 0 0 0 E[YZ] = ? 134 135

Q1a: 10 Points Write the expression for the shaded areas for the following figures: FCAIB.CI AB AT FCAB AB AB MT T'M

Q2b. 12 Points The random variable X takes values -1, 0, 1 with probabilities 1/8, 2/8, 5/8 respectively. Compute E(X) and Var(X). X values I 0 1 Probability I I x2 1 0 1 Ex l 1 E o E i E 3 EX 1 E 101 E 4 5 3 Var X EV EX 4 4 2

Extra Credit. 7 Points John has 2 kids. At least one of them is a boy. What is the probability that both kids are boys? BB BG GB PCB B 3

2/9/2023 2 Topics Chapters 1-3 1. Sets. 2. Counting. 3. Sample space, outcome, event, probability function. 4. Probability: conditional probability, independence, Bayes’ theorem. 5. Discrete/Continuous random variables: events, pmf, cdf. 6. Uniform(n), Poisson & Gaussian random variables 7. E(X), Var(X), σ 8. Transforming random variables (such as Y=aX+b, Y=X^2, Y = f(X), etc.) Midterm Exam: Feb 13, 11-12:30pm ALL students must be on zoom during exam I will take the roll. . 28 29

2/9/2023 3 Example: The graphs below give the PMFs for 3 random variables. Order them by the size of their SD from largest to smallest. (Assume x has the same units in all 3.) C is the largest since: B Is the smallest since A is somewhere in between. Q. 10 Points MV has 2 children. The older child is a girl. What is the probability that both children are girls? Sample Space: {GG, GB} P(GG) = 1/2 31 33

2/9/2023 4 Example: Compute the mean, variance and SD of X. X 1 2 3 4 5 P(x) 1/10 2/10 4/10 2/10 1/10 (X-Mean)^2 4 1 0 1 4 Mean = 1/10 + 4/10 + 12/10 + 8/10 + 5/10 = 3 Variance = 4/10 + 2/10 + 0 + 2/10 + 1/10 = 1.2 SD = SQRT(1.2) Example: 15 Points 1. Suppose random variable X has pdf f(x) = c 𝑥 ଶ 0 ≪ 𝑥 ≪ 2 (a) What is the value of c: ∫ 𝑓 𝑥 𝑑𝑥 ଶ ଴ = ∫ 𝑐𝑥 ଶ 𝑑𝑥 = ଶ ଴ ௖௫ య ଷ ቊ 𝑥 = 2 𝑥 = 0 = 1 c= ଷ ଼ (b) Compute the cdf F(x). F(x) = ∫ 𝑐𝑢 ଶ 𝑑𝑢 = ௫ ଴ ௖ ଷ 𝑥 ଷ = ଵ ଼ 𝑥 ଷ 0 ≪ 𝑥 ≪ 2 (c) Compute P(1 ≤ X ≤ 2). P(1 ≤ X ≤ 2) = F(2) – F(1) = ଼ ଼ − ଵ ଼ = ଻ ଼ 35 37

2/9/2023 5 Example: 15 Points 2. 10 Points Suppose Y has range [0, b] and cdf F(y) = y^2 / 9. (a) What is b? b^2/9 = 1 therefore, b=3 (b) Find the pdf of Y. f(y) = ଶ ଽ y 0 ≪ 𝑦 ≪ 3 Example: 10 Points Which of the following are graphs of valid cumulative distribution functions? NYYN 38 40

2/9/2023 6 Question 16: 20 Points Given the sets L & R, determine : LUR, LꓵR, L’, L-R = L+(-R) = LUR’ 44 Write the expression for the shaded regions. C+AB’ AB+BC 42 44

2/9/2023 7 Problem 5: 10 Points DNA is made of sequences of nucleotides: A, C, G, T. How many DNA sequences of length 3 are there? How many DNA sequences of length 3 are there with no repeats? 4 x 4 x 4 = 64 4 x 3 x 2 = 24 Problem 6: 10 Points There are 5 competitors in a 100m free style swimming final. How many ways can gold, silver and bronze be awarded? 5 x 4 x 3 46 48

2/9/2023 8 . Problem 7: 15 Points A Popular fashion magazine received the following question: Dear Abby, I won’t wear green and red together; I think black or denim goes with any other color in my wardrobe: Here is my wardrobe: Shirts: 3B, 3R, 2G; sweaters 1B, 2R, 1G; pants 2D,2B. (Shirts: 3R) x (Sweater: 1B+2R=3BR) x (Pants: 4BD) = 3x3x4 = 36 (Shirts: 3B) x (Sweater: 4RGB) x (Pants: 4B,D) = 3x4x4 = 48 (Shirts: 2G) x (Sweater: 2BG) x (Pants: 4BD) = 2x2x4= 16 Total: 36 + 48 + 16 = 100 Number: 3 to 4 times 50 52

2/9/2023 9 Problem 8: 10 Points (a) Count the number of ways to get exactly 3 heads in 10 flips of a coin. (b) For a fair coin, what is the probability of exactly 3 heads in 10 flips? (a) Probability (3H, 7T) = ଵ଴ ଷ = 120 combinations (b) There are 2 ଵ଴ possible outcomes for 10 flips. For a fair coin, each outcome is equally probable so the probability of exactly 3 heads is: P(3H,7T) = ଵ଴ ଷ / 2 ଵ଴ = 120/1024 = 0.117 Problem 10: 15 Points You toss a fair coin three 3 times. Write down the event you get 2 or more heads. (HHH, HHT, HTH, THH) Write down the events you get exactly 2 heads. (THH, HTH, HHT) Write down the events you get exactly 1 tail. (THH, HTH, HHT) Are the events “exactly 2 heads” and “exactly 2 tails” disjoint (any overlap)? (THH, HTH, HHT) ꓵ (TTH, THT, HTT) = 0 (empty) Which event is the subset of the other: “at least 2 heads” & “exactly 2 heads”? (THH, HTH, HHT) (THH, HTH, HHT, HHH) 54 56

2/9/2023 10 A class has 50 students; 20 male (M), 25 brown-eyed (B). For a randomly chosen student what is the range of possible values for p = P(M ∪ B)? P(M ∪ B) = P(M) + P(B) − P(M ∩ B) = 2/5 + 1/2 − P(M ∩ B) = 0.9 - P(M ∩ B) The maximum possible value of P(M ∪ B) happens if M and B are disjoint, so P(M ∩ B) = 0. [25Fs are B-eyed] The minimum happens when M ⊂ B, so P(M ∩ B) = P(M) = .4. [B-eyes: 20M, 5F] Therefore, 0.9-0.4 = 0.5 ≤ p ≤ 0.9-0 = 0.9 You have an unfair (p ≠ q) coin with p, probability of a head. You flip the coin twice. What is the probability of getting the same (HH, TT) vs. different (HT, TH). Second Flip H T First Flip H p^2 pq T qp q^2 P(same) = p^2 + q^2 P(diff) = 2pq P(same) > p(diff) 58 60

2/9/2023 11 You have an unfair (p ≠ q) coin with p, probability of a head. You flip the coin twice. What is the probability of getting the same (HH, TT) vs. different (HT, TH). Second Flip H T First Flip H p^2 pq T qp q^2 P(same) = p^2 + q^2 P(diff) = 2pq P(same) > p(diff) Hint: (a-b)^2 > 0 a^2 + b^2 -2ab>0 a^2 + b^2 > 2ab Q. What if p=1? EE 114 Probability, Random Variables, and Random Processes in Electrical Engineering UC Riverside Winter 2023 Instructor: Matt Vaezi Wednesday, February 8, 2023 Review Midterm Exam & Sample Midterm Questions Part II via Zoom Video Conferencing 11-12:30pm ucr.zoom.us Link: https://ucr.zoom.us Meeting ID:910 9686 0815 61 62

2/9/2023 12 Calendar as of Feb 8 Assignment Due Date – Status Reading Assignment Chapters 1-3, CDF & PDF (Chapter 4) HW Set III Solutions will be posted Do not submit Midterm Exam Sample Questions Review – Part II Toss a coin 4 times. Let A = ‘at least three heads’: P(HHHH, THHH, HTHH, HHTH, HHHT) = 5/16 B = ‘first toss is a tail’: P(TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH) = 8/16 P(A ∩ B) = P(THHH) = 1/16 P(A ∩ B) = P(THHH) = 1/16, P(B) = 8/16 P(A|B) = P(A ∩ B) / P(B) = (1/16) / (8/16) = 1/8 P(B|A) = P(A ∩ B) / P(A) = (1/16) / (5/16) = 1/5 63 66

2/9/2023 13 5 red and 2 green balls in an urn. (15 points) A random ball is selected and replaced by a ball of the other color; then a second ball is drawn. 1. What is the probability the second ball is red? 2. What is the probability the first ball was red given the second ball was red? P(R2) = 5/7.4/7 + 2/7.6/7 = 32/49 P(R1|R2) = P(R1 ∩ R2) / P(R2) = (20/49)/(32/49) = 20/32 71 Example: Roll two dice: X = # on first die, Y = # on second die X takes values in 1, 2, . . . , 6, Y takes values in 1, 2, . . . , 6 What is the PMF P(i,j)? Consider the event: A = ‘Y − X ≥ 2’ Describe the event A and find its probability. pmf: p(i, j) = 1/36 for any i and j between 1 and 6. A = {(1, 3),(1, 4),(1, 5),(1, 6),(2, 4),(2, 5),(2, 6),(3, 5),(3, 6),(4, 6)} P(A) = 10 x 1/36 = 10/36 69 71

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2/9/2023 15 Question 1a: 5 Points If you flip a coin 3 times, can you refer to the following event as “Exactly 2 heads”? (THH, HTH, HHT, HHH) True or False? Which of the following describes the event {THH,HTH,HHT}? (1) “exactly one head” (2) “exactly one tail” (3) “at most one tail” (4) none of the above The events “exactly 2 heads” and “exactly 2 tails” are disjoint. True or False If Var(X) = 0 then X is constant. True or False X is constant if and only if Var(X) = 0 True or False Question 1b: 5 Points Given the following figure, what is P(A) in terms of P(B1), P(B2) and P(B3)? Solution: P(A) = P(A ∩ B1) + P(A ∩ B2) + P(A ∩ B3) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) Question 1c: 10 Points Let C and D be two events with P(C) = 0.25, P(D) = 0.45, and P(C ∩ D) = 0.1. What is P(C’∩D)? 75 77

2/9/2023 16 Question 1b: 5 Points Given the following figure, what is P(A) in terms of P(B1), P(B2) and P(B3)? Solution: P(A) = P(A ∩ B1) + P(A ∩ B2) + P(A ∩ B3) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) Question 1c: 10 Points Let C and D be two events with P(C) = 0.25, P(D) = 0.45, and P(C ∩ D) = 0.1. What is P(C’∩D)? Solution: P(D ∩ C) + P(D ∩ C’ ) = P(D) P(D ∩ C’) = P(D) − P(D ∩ C) = 0.45 − 0.1 = 0.35. Question 3: 15 Points Suppose that X is a random variable that takes on values 0, 2 and 3 with probabilities 0.3, 0.1, 0.6, respectively. Let Y = 3(X − 1)^2. (a) What is the expected value of X? (b) What is the variance of X? (c) What is the expected value of Y ? (d) Let FY(t) be the cdf of Y . What is FY(7)? Solution We first make the probability tables X 0 2 3 prob 0.3 0.1 0.6 Y 3 3 12 E(X) = 0 · 0.3 + 2 · 0.1 + 3 · 0.6 = 2 E(X^2 ) = 0 · 0.3 + 4 · 0.1 + 9 · 0.6 = 5.8 Var(X) = E( 𝑥 ଶ ) − E(X)^2 = 5.8 − 4 = 1.8. E(Y ) = 3 · 0.3 + 3 · 0.1 + 12 · 0.6 = 8.4 FY (7) = Prob(Y ≤ 7) = 0.1 + 0.3 = 0.4 78 81

2/9/2023 17 Question 4: 15 Points The events A, B and C are mutually independent with P(A) = 0.3, P(B) = 0.4, P(C) = 0.5. Compute the following: (i) P(A ∩ B ∩ C’) (ii) P(A ∩ B’ ∩ C) (iii) P(A’ ∩ B ∩ C) Solution: By the mutual independence, we have (i) P(A ∩ B ∩ C’) = P(A)P(B)P(C’) = 0.3 x 0.4 x (1-.5) = 0.06 (ii) P(A ∩ B’ ∩ C) = P(A)P(B’)P(C) = 0.3 x (1-0.4) x 0.5 = 0.09 (iii) P(A’ ∩ B ∩ C) = P(A’)P(B)P(C) = (1-0.3) x 0.4 x 0.5 = 0.14 83 87

2/9/2023 18 Example - Solution Given Figure 2.2, let’s assume 𝑝 ௜ is the probability that unit I is operational. What is the probability that the serial and parallel systems are operational? Serial System: P(Components in series being operational) = P(U1 ∩ U2 ∩ U3) = P(U1) P(U2) P(U3) = p1 p2 p3 Parallel System The operation fails when all components fail: P(Failure) = (1-p1)(1-p2)(1-p3) P(Success) = 1-P(Failure) = 1- (1-p1)(1-p2)(1-p3) 88 89

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2/9/2023 20 Given fX(x) shown below, calculate E(x). Take 2 minutes to think about the solution. 93