2005 spring solutions - MATH 218 FINAL EXAM SOLUTIONS MAY 3...

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MATH 218 FINAL EXAM SOLUTIONS MAY 3, 2005 Solutions prepared by Ronald Bruck, May 2005. N ote : Answers have been marked in red. The instructions specifically noted that answers were to be given either as exact fractions reduced to lowest terms (when that was possible), or (if given numerically) to four exact digits after the decimal point. Also, answers were to be placed in answer boxes, where provided. It is a very fortunate thing for some students that the examiners didn’t insist on these conditions. They will not always be so lucky. A remark on calculation: to compute to “four exact digits” requires comput- ing to five digits. The extra digit is to prevent roundoff error, and is called a “guard” digit. What does not work is to round off 4 / 3 to 1 . 3, use 1 . 3 through- out a calculation, and then try to get the answer to four decimals! Once you’re rounded off, you’ve lost that extra precision forever. Rounding 4 / 3 to 1 . 3 re- sults in two significant figures of precision, at most. Problem 1 (20 points) . Your company receives widgets from 3 suppliers, A, B and C. Sixty percent of the widgets come from A, thirty percent come from B, and the rest come from C. Company A is excellent, and only 5 percent of their widgets are defective. Company B is not so good, and 20 percent of their widgets are defective. Company C is pretty bad, and half of their widgets are defective. You take a random widget arriving at your company and test it. Write D for the event that it is defective. 1a) Draw a tree diagram to represent this situation. Include all events and probabilities involved—individual, conditional, and joint (inter- section) probabilities. [ REMINDERS: branches from the root are labeled with individual probabilities, further branches are labeled with conditional probabilities, and terminal nodes represent inter- sections and should be labeled with the (joint() probability of that intersection]. 1b) What is the chance that the tested widget is defective? 1c) Suppose the tested widget is defective. What is the chance that it came from A? 1d) Suppose the tested widget is defective. What is the chance that it came from B? 1e) Suppose the tested widget is defective. What is the chance that it came from C? 1f) Suppose the tested widget is NOT defective. What is the chance that it came from A? 1g) [Yes or no.] Should the probabilities for 1c), 1d), and 1e) add up to 1? 1h) [Yes or no.] Should the probabilities for 1c), 1f) add up to 1? Solution. 1a) The tree is in Figure (1), on the next page. A D D 0.05 0.95 P(A&D)=0.6x0.05=0.03 P(A&D)=0.6x0.95=0.57 B D D 0.20 0.80 P(B&D)=0.3x0.2=0.06 P(B&D)=0.3x0.8=0.24 C 0.6 0.3 0.1 D D 0.50 0.50 P(C&D)=0.1x0.5=0.05 P(C&D)=0.1x0.5=0.05 F igure 1. Tree diagram for widget suppliers (A, B, C) vs. defective (D) 1b) D = ( D & A ) ( D & B ) ( D & C ) , and these are mutually exclusive and have probabilities 0 . 03, 0 . 06, and 0 . 05, thus P ( D ) = 0 . 03 + 0 . 06 + 0 . 05 = 0 . 14 .
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