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ProbStats HWSolutions

# ProbStats HWSolutions - Solutions to Homework Problems...

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Solutions to Homework Problems, 550.311 Spring 2008 Collected and Solutions by Donniell Fishkind. Many examples are taken from standard textbooks by Jay Devore, Neil Weiss, Bernard Rosner, John Rice, Sheldon Ross, Cincich, Levine & Stephan, Walpole, Myers, Myers, & Ye. Probability basics Problem 1.1: Later in the semester’s homework you will show that if a 1 inch needle is ran- domly dropped on a floor lined with parallel lines spaced 2 inches apart, then the probability of the needle touching a line is exactly 1 π . Assuming this is true, assuming the only arithmetic operation you may do is to count, and assuming you have access to such a needle and such a floor, how might you compute an integer numerator and an integer denominator of a fraction approximating π ? Solution: Drop the needle many, many times, say n >> 0. If the number of those times that the needle touches a line is x , then we know that x n 1 π , thus n x π . Problem 1.2: (Devore ed5, p58) An engineering construction firm is currently working on power plants at three different sites. Let A i denote the event that the plant at site i is completed by the contract date. Use the operations of union, intersection, and complementation to describe each of the following events in terms of A 1 , A 2 , and A 3 , draw a Venn diagram, and shade the region corresponding to each one: a) At least one plant is completed by the contract date. b) All plants are completed by the contract date. c) Only the plant at site 1 is completed by the contract date. d) Exactly one plant is completed by the contract date. e) Either the plant at site 1 or both of the other two plants are completed by the contract date. Problem 1.3: (Rice p25) Show from the probability axioms: For any events A , B it holds that P( A B ) P( A )+P( B ) 1. (You may use the results derived in class without reproving them.) Solution: In class, we derived from the axioms that P( A B ) = P( A ) + P( B ) P( A B ), thus P( A B ) = P( A ) + P( B ) P( A B ) P( A ) + P( B ) 1, the last inequality holding since P( A B ) 1, i.e. P( A B ) ≥ − 1. Problem 1.4: Show from the probability axioms: For any events E 1 , E 2 , E 3 , . . . (even if they are 1

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not disjoint) it holds that P( i E i ) i P( E i ). [Hint: Consider events of form A i := E i \ ( i 1 j =1 E j ).] Solution: Following the hint, we consider events A i := E i \ ( i 1 j =1 E j ) for i = 1 , 2 , 3 , . . . . Note that A 1 , A 2 , A 3 , . . . are disjoint; if an outcome is in A k then it is in E k , hence not in A m for any k, m such that k < m . Consequently, by the axioms, P( i A i ) = i P( A i ). Also note that i A i = i E i , and note that for all i A i E i hence P( A i ) P( E i ). Thus, P( i E i ) = P( i A i ) = i P( A i ) i P( E i ). Combinatorial probability Problem 2.1: (Devore ed5, p74) A class has 20 nonsmokers, 15 light smokers, and 10 heavy smokers. Six of these will be randomly selected to participate in a study (each set of 6 is equiprob- able). a) What is the probability that all 6 are heavy smokers? b) What is the probability that all 6 have the same status (non/light/heavy smoking)?
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