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hw2 - MS&E 120 Probabilistic Analysis Fall 2007-2008 Prof...

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MS&E 120: Probabilistic Analysis Fall 2007-2008, Prof. Shachter HOMEWORK #2 SOLUTIONS CHAPTER #2 PROBLEMS 12) a) Using Proposition 4.4, we have P(student in Spanish class U student in French class U student in German class) =1/100 x [(28+26+16) – (12+4+6) + 2] = 0.5 P(student not taking any of these classes) = 1 – 0.5 = 0.5 b) # students taking exactly Spanish = 28 – (12+4) + 2 = 14 # students taking exactly French = 26 – (12+6) + 2 = 10 # students taking exactly German = 16 – (4+6) + 2 = 8 P(students taking exactly one class) = 1/100 x (14+10+8) =0.32 c) P(at least one student is taking a language class) = 1 – P(neither student is taking a language class) = 1 – (50/100 x 49/99) = 0.7525 15) a) First we pick 1 suit out of 4, and then 5 ranks out of the 13 ranks in the suit. # ways to make a flush = 5 13 1 4 Hence P(flush) = 5 52 5 13 1 4 = 0.00198 b) We first determine the number of ways to make a one-pair 5 card hand. This is equivalent to choosing one of the 13 ranks (Ace to King) in a standard deck, choosing two suits with these ranks, choosing 3 different ranks of the remaining 12, and of each of those 3 ranks choosing one suit. Thus # of ways to make a one-pair hand = 1 4 1 4 1 4 3 12 2 4 1 13 = 1098240
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Thus P(one pair) = 5 52 1098240 = 0.4226 c) We have a similar situation here, except we choose two distinct ranks from 13, from which we choose 2 out of 4 suits for the first rank and for the second, and finally, choose one from the final 44 cards left over. # of ways to make two-pair hand = 1 4 1 11 2 4 2 4 2 13 = 123552 Thus P(two pair) = 5 52 123552 = 0.0475 d) By the same reasoning as above, # ways to make 3 of a kind = 1 4 1 4 2 12 3 4 1 13 = 54912 Thus P(3 of a kind) = 5 52 54912 = 0.0211 e) # ways to make 4 of a kind = 1 4 1 12 4 4 1 13 = 624 Thus P(4 of a kind) = 5 52 624 = 0.00024 21) a) P(#children = 1) = 4/20 = 0.2 P(#children = 2) = 8/20 = 0.4 P(#children = 3 ) = 5/20 = 0.25 P(#children = 4 ) = 2/20 = 0.1 P(#children = 5) = 1/20 = 0.05 Total # of children = 4+16+15+8+5=48 P(child from family with 1 child) = 4/48=0.08333 P(child from family with 2 children) = 16/48=0.3333 P(child from family with 3 children) = 15/48=0.3125 P(child from family with 4 children) = 8/48=0.1666
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P(child from family with 5 children) = 5/48=0.1042 =0.48 43) a) Since A and B have to stick together, we can treat A and B as one person. Therefore we have (N-1) people to permute without restrictions. Moreover, we can permute A and B among themselves. So there are (N-1)! 2! ways for
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