Tutorial 9 Solutions

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University of Toronto Department of Mechanical and Industrial Engineering MIE236: Probability (Fall 2022) Tutorial 9: Review Problem 1: A producer of a certain type of electronic component ships to suppliers in lots of twenty. Suppose that 60% of all such lots contain no defective components, 30% contain one defective component, and 10% contain two defective components. A lot is picked, two components from the lot are randomly selected, and neither is defective. a) What is the probability that zero defective components exist in the lot? b) What is the probability that only one defective exists in the lot? c) What is the probability that two defectives exist in the lot? Solution: Consider the following events: ?: ? ??? ???? ??? ?????𝑖? ??? ??????𝑖?? ?????????? ?: ? ??? ?????𝑖?? ??? ??????𝑖?? ????????? ?: ? ??? ???????? ??? ??????𝑖?? ?????????? ?: ???? ?? ??? ???????? ?????????? ??? ????????? We know that 𝑷(?) = ?. ? , 𝑷(?) = ?. ? , and 𝑷(?) = ?. ? ; and we can compute the conditional probabilities 𝑷(?|?) = ? , 𝑷(?|?) = ( ?? ? ) ( ?? ? ) ⁄ = ?. ? , and 𝑷(?|?) = ( ?? ? ) ( ?? ? ) ⁄ = ?. ??? . Then, we can compute: a) 𝑷(?|?) = 𝑷(?|?)𝑷(?) 𝑷(?) = 𝑷(?|?)𝑷(?) 𝑷(?|?)𝑷(?)+𝑷(?|?)𝑷(?)+𝑷(?|?)𝑷(?) = ?.? ?.?+?.?⋅?.?+?.???⋅?.? = ?. ???? b) 𝑷(?|?) = 𝑷(?|?)𝑷(?) 𝑷(?) = ?.?⋅?.? ?.?+?.?⋅?.?+?.???⋅?.? = ?. ???? c) 𝑷(?|?) = ? − 𝑷(?|?) − 𝑷(?|?) = ?. ???? Problem 2: A nationwide survey of young men in their 20s revealed that 70% disapprove of buying cryptocurrency, according to a report from Chinese government. If 12 men in their 20s are selected at random and asked their opinion, find the probability that the number who disapprove of buying cryptocurrency is: a) anywhere from 7 to 9; b) at most 5; c) not less than 8.
University of Toronto Department of Mechanical and Industrial Engineering MIE236: Probability (Fall 2022) Solution: From the look-up Table A.1 with n = 12 and p = 0.7, we have: (a) 𝑷(? ≤ 𝑿 ≤ ?) = 𝑷(𝑿 ≤ ?) − 𝑷(𝑿 ≤ ?) = ?. ???? − ?. ???? = ?. ???? (b) 𝑷(𝑿 ≤ ?) = ?. ???? (c) 𝑷(𝑿 ≥ ?) = ? − 𝑷(𝑿 ≤ ?) = ? − ?. ???? = ?. ???? Problem 3: An engineering student club lists as its members: 2 mechanical engineering students, 3 civil engineering students, 5 computer science students, and 2 chemical engineering students. If a representative group of 4 is selected at random to attend a national-wide competition, find the probability that: a) all four programs are represented; d) all programs except computer science students are represented. Solution: a) Using the extension of the hypergeometric distribution, we have: ( ? ? )( ? ? )( ? ? )( ? ? ) ( ?? ? ) = ? ?? b) Using the extension of the hypergeometric distribution, we have: ( ? ? )( ? ? )( ? ? ) ( ?? ? ) + ( ? ? )( ? ? )( ? ? ) ( ?? ? ) + ( ? ? )( ? ? )( ? ? ) ( ?? ? ) = ? ??? Problem 4: A current of ? amperes flowing through a resistance of 𝑅 ohms varies according to the probability distribution ?(𝑖) = { 6𝑖(1 − 𝑖), 0 < 𝑖 < 1, 0, ?????ℎ???. If the resistance varies independently of the current according to the probability distribution ?(?) = { 2?, 0 < ? < 1, 0, ?????ℎ???, Find the probability distribution for the power ? = ? 2 𝑅 watts.
University of Toronto Department of Mechanical and Industrial Engineering MIE236: Probability (Fall 2022) Solution: We use the fact that ? and 𝑅 are independent to get the joint probability distribution: ?(𝑖, ?) = 12?𝑖(1 − 𝑖), 0 < 𝑖, ? < 1. Let ? = ? 2 𝑅 and ? = 𝑅 be the transformed variables we care about. The inverse transformations are ? = ? and 𝑖 = √?/? , for 0 < ? < 1 and ? < ? < 1. We compute the Jacobian as follows: ? = | 𝜕𝑖 𝜕? 𝜕𝑖 𝜕? 𝜕? 𝜕? 𝜕? 𝜕? | = | 1 2 √ ?? − √ ? 2? √ ? 0 1 | = 1 2 √ ?? Which gives us the joint probability distribution of ? and ? : ?(?, ?) = ?(?, ?)|?| = 12?√?/? (1 − √?/? ) 1 2 √ ?? = 6(1 − √?/? ) Finally, we compute the marginal distribution of ? : ℎ(?) = ∫ 6(1 − √?/? ) ?? 1 ? = 6(? − 2 √ ?? | ? 1 = 6 + 6? − 12 √ ? . Problem 5: Let ? be a continuous random variable with the probability density function ?(?) = { 2/? 2 if 1 < ? < 2 0 elsewhere Find the distribution and the density functions of ? = ? 2 . Solution: Let 𝐺 and ? be the distribution function and the density function of ? , respectively. By definition, 𝐺(?) = 𝑃(? ≤ ?) = 𝑃(? 2 ≤ ?) = 𝑃(− √ ? ≤ ? ≤ √ ? ) = { 0 ? < 1 𝑃(1 ≤ ? ≤ √ ? ) 1 ≤ ? < 4 1 ? ≥ 4. Now 𝑃(1 ≤ ? ≤ √ ? ) = ∫ √𝑡 1 2 ? 2 ?? = [− 2 ? ] 1 √𝑡 = 2 − 2 √? Therefore,
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