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Homework #3. Due: Friday, September 12, 2003. <KEY> IE 230 Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John Wiley & Sons, New York, 2003. Chapter 2, Sections 2.3–2.5. 1. 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 Ai 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 2) = 0.88. (a) In words, what is (A 1 ∩ A 2)? _____________________________________________________________ Both passengers arrive to claim a seat. _____________________________________________________________ (b) Determine the probability that at least one passenger arrives to claim a seat. _____________________________________________________________ P(A 1 ∪ A 2) = P(A 1) + P(A 2) − P(A 1 ∩ A 1) = 0.9 + 0.95 − 0.88 = 0.97 ← Always true Substitute given values Simplify (Comment: In the original version of this problem, P(A 1 ∩ A 2) = 0.8 was given, leading to a probability greater than one. Think about why that probability value is impossible. _____________________________________________________________ – 1 of 5 – Schmeiser Homework #3. Due: Friday, September 12, 2003. <KEY> IE 230 (c) Determine the probability that neither passenger arrives to claim a seat. _____________________________________________________________ P[(A 1 ∪ A 2)′] = 1 − P(A 1 ∪ A 2) = 1 − 0.97 = 0.03 ← Complement From Part (b) 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. In A 1 ∩ A 2, write 0.88. In A 1 ∩ A 2′, write 0.02. In A 1′ ∩ A 2, write 0.07. In A 1′ ∩ A 2′ = (A 1 ∪ A 2)′, write 0.03. (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 1 ∩ A 2) / P(A 1) = 0.88 / 0.90 = 0.987 ← P(A 2 | A 1) Def. of conditional prob. Substitute known values Simplify (Comment: Can you see the answer in the Venn Diagram of Part (d)?) _____________________________________________________________ (f) Determine P(A 2 | A 1′). _____________________________________________________________ P(A 2 | A 1′) = = = = P(A 1′ ∩ A 2) / P(A 1′) P(A 1′ ∩ A 2) / [1 − P(A 1)] 0.07 / (1 − 0.90) 0.7 ← Def. of conditional prob. Complement Substitute known values Simplify (Comment: Can you see the answer in the Venn Diagram of Part (d)?) _____________________________________________________________ (g) Determine P(A 2′ | A 1). _____________________________________________________________ P(A 2′ | A 1) = P(A 1 ∩ A 2′) / P(A 1) = 0.02 / 0.90 = 0.222 ← Def. of conditional prob. Substitute known values Simplify (Comment: Can you see the answer in the Venn Diagram of Part (d)?) _____________________________________________________________ – 2 of 5 – Schmeiser Homework #3. Due: Friday, September 12, 2003. <KEY> IE 230 (h) Determine P(A 2′ | A 1′). _____________________________________________________________ P(A 2′ | A 1′) = = = = = P(A 1′ ∩ A 2′) / P(A 1′) P[(A 1 ∪ A 2)′] / P(A 1′) [1 − P(A 1 ∪ A 2)] / [1 − P(A 1)] (1 − 0.97) / (1 − 0.90) 0.3 ← Def. of conditional prob. DeMorgan’s Law Complements Substitute known values Simplify (Comment: Can you see the answer in the Venn Diagram of Part (d)?) _____________________________________________________________ (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 = 0.375 ← _____________________________________________________________ (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 = 0.359 ← _____________________________________________________________ – 3 of 5 – Schmeiser Homework #3. Due: Friday, September 12, 2003. <KEY> IE 230 (e) Determine P(B ∩ A ). _____________________________________________________________ P(B ∩ A ) = P(A )P(B | A ) = (15 / 40)(14 / 39) = 0.135 ← (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) − 0.135 = 0.615 ←. (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) = 0.375 = 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′ ) = (0.02)(0.7) + (0.03)(0.3) = 0.023 ← Total Probability Substitute known values _____________________________________________________________ (e) Determine the probability that a randomly selected roll does not contain flaws. _____________________________________________________________ P(F ) = 1 − P(F ) = 1 − 0.023 = 0.097 ← _____________________________________________________________ – 4 of 5 – Schmeiser Homework #3. Due: Friday, September 12, 2003. <KEY> IE 230 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′ ) = (1)(1 / 4) + (0)(3 / 4) = 1/4← Total Probability Assume equally likely choices simplify _____________________________________________________________ (b) If the contestant does switch doors. _____________________________________________________________ P(W ) = P(W | C )P(C ) + P(W | C′ )P(C′ ) = (0)(1 / 4) + (1)(3 / 4) = 3/4← Total Probability Assume equally likely choices simplify _____________________________________________________________ – 5 of 5 – Schmeiser