1. Two useful probability results. Provide the reason for each step of the proofs. (a) Result: P(A ) + P(A ) = 1. = P(A ) + P(A ) < Axiom 3 >

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1 Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John Wiley & Sons, New York, Chapter 2, Sections Two useful probability results. Provide the reason for each step of the proofs. (a) Result: P(A ) + P(A ) = 1. Proof. 1 = P(S ) < Axiom 1 > = P(A A ) < Same event > = P(A ) + P(A ) < Axiom 3 > (b) Result: P(A B ) = P(A ) + P(B ) P(A B ) Proof. P(A B ) = P[(A B ) (A B ) (A B )] < Same event > = P[(A B ) (A B )] + P(A B ) < Axiom 3 > = P(A B ) + P(A B ) + P(A B ) < Axiom 3 > = [P(A B ) + P(A B )] + [P(A B ) + P(A B )] P(A B ) < Algebra > = [P(A B ) (A B )] + [P(A B ) (A B )] P(A B ) < Axiom 3 > = P(A ) + P(B ) P(A B ) < Same events > 2. Consider two passengers with tickets for a particular air flight. Let A i denote the event that the i th passenger arrives to claim a seat. Suppose that P(A 1 ) = 0.9, P(A 2 ) =.95 and P(A 1 ) = (a) In words, what is (A 1 )? Both passengers arrive to claim a seat. (b) Determine the probability that at least one passenger arrives to claim a seat. P(A 1 ) + P(A 2 ) P(A 1 A 1 ) Always true = Substitute given values = 0.97 Simplify (Comment: In the original version of this problem, P(A 1 ) = 0.8 was given, leading to a probability greater than one. Think about why that probability value is impossible. 1 of 5 Schmeiser

2 (c) Determine the probability that neither passenger arrives to claim a seat. P[(A 1 ) ] = 1 P(A 1 ) Complement = From Part (b) = 0.03 Simplify (d) Draw a Venn diagram, showing in each of the four areas the corresponding probability. Draw two overlapping circles in a rectangle. Label them A 1 and A 2., write 0.88., write 0.02., write =(A 1 ), write (Comment: Can you argue carefully for each of these four values?) (Comment: Can you redraw the Venn diagram so that the four areas are proportional to the probabilities?) (e) Determine P(A 2 A 1 ). P(A 2 A 1 ) / P(A 1 ) Def. of conditional prob. = 0.88 / 0.90 Substitute known values = Simplify (f) Determine P(A 2 A 1 ). P(A 2 A 1 ) / P(A 1 ) Def. of conditional prob. = P(A 1 ) / [1 P(A 1 )] Complement = 0.07 / (1 0.90) Substitute known values = 0.7 Simplify (g) Determine P(A 2 ). P(A 2 ) / P(A 1 ) Def. of conditional prob. = 0.02 / 0.90 Substitute known values = Simplify 2 of 5 Schmeiser

3 (h) Determine P(A 2 ). P(A 2 ) / P(A 1 ) Def. of conditional prob. = P[(A 1 ) ] / P(A 1 ) DeMorgan s Law = [1 P(A 1 )] / [1 P(A 1 )] Complements = (1 0.97) / (1 0.90) Substitute known values = 0.3 Simplify (i) Explain why P(A 2 A 1 ) and P(A 2 A 1 ) do not need to sum to one. Events are complementary only when discussed under the same conditioning event. Here the conditions are complementary. 3. (Montgomery and Runger, Problem 2 63) A lot contains 15 castings from a local supplier and 25 castings from a supplier in the next state. Two castings are selected at random, without replacement, from the lot of 40 castings. Let A be the event that the first casting is selected from the local supplier and let B be the event that that second casting is selected from the local supplier. (a) What is the procedure that defines the experiment? Select one casting from the total lot of 40. (b) What sample space do you choose to use? In general, we could use the set of all pairs of castings. (Comment: There are (40)(39) / 2 = 780 such pairs. For Parts (c) and (d), however, it is simpler to use smaller sample spaces specific to each part.) (c) Determine P(A ). Let the sample space be the set of forty castings. Then because each casting is equally likely to be chosen, P(A ) = 15 / 40 = (d) Determine P(B A ). Let the sample space be the set of the remaining 39 castings. Then because each casting is equally likely to be chosen, P(A ) = 14 / 39 = of 5 Schmeiser

4 (e) Determine P(B A ). P(B A ) = P(A )P(B A ) = (15 / 40)(14 / 39) = (Comment: We could have used the larger sample space with 780 outcomes and counted the number of pairs for which both are local. Evidently there are 105 such pairs.) (f) Determine P(B A ). P(B A ) = P(B ) + P(A ) (B A ) = (0.375) + (0.375) = (Comment: Is it obvious that P(B ) = P(A )? If not, use Total Probability to find P(B ) = (B A )P(A ) + (B A )P(A ) = (14 / 39)(15 / 40) + (15 / 39)(25 / 40) = = P(A ).) 4. (Montgomery and Runger, Problem 2 73) Suppose that 2% of cotton fabric rolls and 3% of nylon fabric rolls contain flaws. Of the rolls used by a manufacturer, 70% are cotton and 30% are nylon. (a) What procedure defines the experiment that underlies this problem? Randomly choose one of the fabric rolls. (b) Define notation for selecting a cotton roll, for selecting a nylon roll, and for selecting a roll that contains flaws. Let C = "the roll is cotton." Let F = "the roll contains flaws." (c) Write the given information in terms of your notation. P(C ) = 0.7 (and therefore P(C ) = 0.3) P(F C ) = 0.02 P(F C ) = 0.03 (Comment: We could define N = "the roll is nylon", but C suffices.) (d) Determine the probability that a randomly selected roll contains flaws. P(F ) = P(F C )P(C ) + P(F C )P(C ) Total Probability = (0.02)(0.7) + (0.03)(0.3) Substitute known values = (e) Determine the probability that a randomly selected roll does not contain flaws. P(F ) = 1 P(F ) = = of 5 Schmeiser

5 5. Consider the Monte Hall problem. Determine the probability of winning if the game has four doors, there is only one prize, and Monte Hall opens two empty doors. (a) If the contestant does not switch doors. As in class, let W = "the contestant wins" and let C = "the contestant s first choice is correct. P(W ) = P(W C )P(C ) + P(W C )P(C ) Total Probability = (1)(1 / 4) + (0)(3 / 4) Assume equally likely choices = 1 / 4 simplify (b) If the contestant does switch doors. P(W ) = P(W C )P(C ) + P(W C )P(C ) Total Probability = (0)(1 / 4) + (1)(3 / 4) Assume equally likely choices = 3 / 4 simplify 5 of 5 Schmeiser