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# hw1_soln - CS 3341 HW 1 SOLUTION 1 Sec 1.3 1(a Outcome =...

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Page 1 of 6 CS 3341 HW # 1 SOLUTION 1. Sec 1.3, # 1 (a) Outcome = 3-tuple representing the type of the selected (chip 1, chip 2, chip 3). Total # outcomes in S = (2) (2) (2) = 8. S = { (RAM,RAM,RAM), (RAM,RAM,ROM), (RAM,ROM,RAM), (ROM,RAM,RAM), (RAM,ROM,ROM), (ROM,RAM,ROM), (ROM,ROM,RAM), (ROM,ROM,ROM) } (b) Outcome = 3-tuple representing the type of the selected (chip 1, chip 2, chip 3), while keeping in mind that there is only one defective (D) and nine good (G) chips. S = { (G, G, G), (D, G, G), (G, D, G), (G, G, D) } 2. Sec 1.5, # 1 Keep in mind that there are 4 components. (a) B C = {exactly 2 G’s or at most 3 D’s} (b) B C ={exactly 2 G’s and at most 3 D’s} ={exactly 2 G’s and exactly 2 D’s} (c) A C = {all 4 D’s or at most 3 D’s} = {at most 4 D ’s } (d) A C ={} = empty or null event (event not possible) 3. Sec 1.8, # 1 We want two-digit even numbers. Case 1: Assume repetitions allowed. # of choices for the left-most digit = 5. # of choices for the right-most digit = 2 So, total # of choices = 5 (2) = 10 Case 2: Assume repetitions not allowed. 44 and 66 are not possible now. So, total # of choices = 10 2 = 8. 4. Sec 1.8, # 4 Outcome = # defective chips in three selected three chips. Note that the order of selection is not important here since we are only interested in counting the # of defectives.

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Page 2 of 6 Total # of outcomes in S = # of ways to draw 3 chips from 15 chips = “15 choose 3” = 455 . Total # of favorable outcomes = # of ways to draw 3 chips from 5 D chips = “5 choose 3” = 10 So, P(All three D’s) = 10/455 = 0. 022 5. Sec 1.8, # 5 Using the formula derived in the class, P(No match) = 0.973 P(At least one match) = 1 0.973 = 0.027. 6. When a computer goes down, there is a 75% chance that it is due to an overload and a 15% chance that it is due to a software problem. There is an 85% chance that it is due to an overload or a software problem. What is the probability that both of these problems are at fault? What is the probability that
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