ee250_midterm_Spring2024_solutions

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EE250: Probability, Random Variables and Stochastic Processes Spring 2024 Midterm 1 Solutions March 04 th , 2024, 3:00pm-4:15pm Name: ID#: ________________________ This is a close-book, close-notes exam. For this exam, you may use a calculator and two page (8.5*11 inches) double-sided cheat-sheet . All other electronic devices (e.g., cellular phones, laptops, ipads, PDAs, etc.) are prohibited. If you have a cellular phone, please turn it off prior to the start of the exam to avoid distracting other students. The exam contains 3 questions for a total of 100 points. Please answer the questions in the spaces provided. If you need additional space, use the last two blank scratch papers and clearly indicate that you have done so. You will have 75 minutes to complete this exam.

1. (30 points) An integrated circuit fabory has three machines X, Y, and Z. Test one integrated circuit produced by each machine. Either a circuit is acceptable (a) or it fails (f). An observation is a sequence of three test results corresponding to the circuits from machines X, Y, and Z, respectively. For example, aaf is the observation that the circuits from X and Y pass the test and the circuit from Z fails the test. Suppose that the three tested circuits are independent and the probability of an acceptable circuit is 0.8. a. What are the elements of the sample space of this experiment? b. What are the elements of the sets C = {more than two circuits acceptable}, D = {at least one circuit fail}. c. Are C and D independent? Explain your reasoning. d. Find the probability P[ ? 𝐶 ∪? 𝐶 ] a. S={aaa, aaf, faa, afa, aff, faf, ffa, fff} 8 outcomes b. C={aaa} D={faa,afa,aaf,ffa,faf,aff, fff} c. P[C]=0.8^3, P[D]=0.2*0.8^2*3+0.2^2*0.8*3+0.2^3 P[C ∩ D]=P[ Φ ]=0 ≠ P[C]*P[D] Not independent d. P[ ? 𝐶 ∪? 𝐶 ]=P[ (? ∩ ?) ? ]=1-P[ ? ∩ ? ]=1

2. (35 points) Monitor four consecutive packets going through an Internet router. Classify each one as either video (v) or data (d). Your observation is a sequence of four letters (each one is either v or d). For example, four video packets corresponds to vvvv. The outcomes vvvv and dddd each have probability 0.1 whereas each of the other outcomes has the same probability. Count the number of video packets Nv in the four packets you have observed. Calculate the following probabilities: (a) P[Nv=2] (b) P[Nv≥1] (c) P[{vdvd, vvvd}|Nv=2] (d ) P[Nv=2| Nv≥1] (e) P[Nv≥1| Nv=2] S={vvvv, dddd, vddd, dvdd, ddvd, dddv, dvdv, vdvd, vvdd, ddvv, dvvd, vddv, vvvd, vvdv, vdvv, dvvv} P[vvvv]=0.1 P[dddd]=0.1 P[each other outcome]=(1-0.1-0.1)/14=0.8/14 (a) {Nv=2}={ dvdv, vdvd, vvdd, ddvv, dvvd, vddv } P[Nv=2]=0.8/14*6 (b) { Nv≥1 }={ vvvv, vddd, dvdd, ddvd, dddv, dvdv, vdvd, vvdd, ddvv, dvvd, vddv, vvvd, vvdv, vdvv, dvvv } P[ Nv≥1 ]=0.1+0.8/14*14=0.9 (c) P[{vdvd, vvvd}|Nv=2]= 𝑃[{vdvd,vvvd}∩{𝑁𝑣=2}] 𝑃[𝑁𝑣=2] = 𝑃[𝑣?𝑣?] 𝑃[𝑁𝑣=2] = 0.8 14 0.8 14 ∗6 = 1 6 {Nv ≥ 1} ∩ {𝑁𝑣 = 2} = {𝑁𝑣 = 2} (d) P[Nv=2| Nv≥1] = 𝑃[{Nv≥1}∩{𝑁𝑣=2}] 𝑃[𝑁𝑣≥1] = 𝑃[Nv=2] 𝑃[𝑁𝑣≥1] = 0.8 14 ∗6 0.9 (e) P[Nv≥1| Nv=2] = 𝑃[{Nv≥1}∩{𝑁𝑣=2}] 𝑃[𝑁𝑣=2] = 𝑃[Nv=2] 𝑃[𝑁𝑣=2] = 1

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3. (35 points)A box of 23 memory chips has 10 functioning chips, 5 defective but repairable chips, and 8 kaput chips. a) Suppose we select 11 chips from the box, without replacement, with order . Let event A be: there are 5 functioning chips selected in the sample. Find P[A]. b) Suppose we select 11 chips from the box, with replacement, with order . Let event B be: there are 5 functioning chips. Find P[B]. c) Suppose we select 5 chips from the box, with replacement, without order . The box is acceptable if at least 4 of the selected chips are found functioning; otherwise, the box is rejected. Find the probability that the box is acceptable. a) number of outcomes in the sample space is 23! (23−11)! Number of outcomes in event A is 10! (10−5)! 13! (13−6)! ( 11 5 ) P[A]= 10! (10−5)! 13! (13−6)! ( 11 5 ) 23! (23−11)! b) number of outcomes in the sample space is 23 11 Number of outcomes in event B is 10 5 13 6 ( 11 5 ) P[B]= 10 5 13 6 ( 11 5 ) 23 11 c) P[the box is acceptable]=P[4 of the selected chips are functioning]+ P[5 of the selected chips are functioning]= ( 10−1+4 10−1 )( 13−1+1 13−1 ) ( 23−1+5 23−1 ) + ( 10−1+5 10−1 ) ( 23−1+5 23−1 )