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1. Your special Valentine is late mailing you a box of chocolates. You estimate that there is a 75% chance that SX shipping will be chosen and a 25% chance that UVS shipping will be chosen (because it is more expensive). From your experience with them, SX has about a 24% chance of getting the box to you on time if it is chosen; while UVS, which is faster, has about a 45% chance of getting you the box on time if it is chosen. a) (10 pts.) What is the probability the candy will arrive on time in an SX package? P(Timely,SX) = P(SX)P(Timely|SX) = .75x.24 = .18 What is the probability the candy will arrive on time in a UVS parcel? P(Timely,UVS) = P(UVS)P(Timely|UVS) = .25x.45 = .1125 Therefore, what is the probability that your Valentine chocolates will not arrive on time? P(Timely) = P(Timely,SX) + P(Timely,UVS) = .18 + .1125 = .2925 P(Late) = 1 P(Timely) = .7075 b) (10pts) Valentines day comes to an end, and no candy has arrived. What is the probability that this is because your sweetheart chose SX shipping? P(SX|Late) = P(Late,SX)/P(Late) = P(SX)P(Late|SX)/P(Late) = .75x(1.24)/ 7075 =.8057 2. At a Valentine's Day dance, the first and last dances are with your date, of course. But for each of the other 6 dances, you will be blindfolded, and shoot an arrow at a board where pictures of the 20 possible dance partners are posted, spaced out evenly. You will dance with the partner whose picture is closest to where your arrow hits on the board. a) (10 pts.) What is the probability that none of the 6 dances will be with your date? P not date ! 6 from 20 = # 6 from 19 # 6 from 20 = 19 6 = .735 20 6 b) (10 pts.) What is the probability that, during the 6 dances, you will have 5 or fewer different dance partners? P 5 or fewer ! 6 from 20 = 1 - P 6 different ! 6 from 20 = 20 6 1- 6 = 1 - .436 = .564 20 3. Given any two events A and B (and a condition C that is assumed true throughout this problem) plus any rule for calculating probabilities, it turns out to always be true that (in shorthand, which does not mention the condition C) P A - B = P A + P B - A - P B . a) (10 pts.) Explain why this is true, using only the four Axioms of Probability. Do not use counting or areas in your explanation. If you draw pictures, they will only be helpful illustrations, and not part of the answer. (Hint: it will turn that you will need only one of the four axioms, but you will use it twice.) 2 The only axiom that mentions set subtraction is #3, which says that (in shorthand) for any sets D and E, we have P DE = P D + P E - D . That can be used in two ways, for P AB = P A + P B - A each of our two set subtractions: and P BA = P B + P A - B . But the two left sides mean the same thing, so the two right sides are equal: P A + P B - A = P B + P A - B . Moving P(B) to the other side of the equation gets our equation. b) (5 pts.) During the recent epidemic at Tech, the probability that someone would catch a cough was .32, and the probability that someone would show a fever was .19. We learn further that the probability someone would show a fever but not a cough was .10. What is the probability that someone would catch a cough, but not show a fever? P(CF) = P(C) + P(FC ) P(F) = .32 + .10 .19 = .23 4. Amy, Beth, and Carol decide to split up the 13 chocolate truffles in a box among them, by letting each candy in turn be equally likely to go to each of them by chance. They do not try for equal numbers (so that by pure but amazing luck, Carol might get all the candy). a) (5 pts.) In how many ways could the thirteen chocolates be allocated? 13 Each chocolate can go to one of 3 people so allocations = 3 = 1,594,323 . b) (5 pts.) In how many ways could 5 go to Amy, 4 to Beth, and 4 to Carol? 13 = 90,090 544 c) (5 pts.) What is the probability of the outcome in b), if all possible allocations in a) are equally likely? P b = 90090 = .0565 1594323 ... View Full Document

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