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# 1.3 - 10 and p n = 0 for all other numbers If q n is the...

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Solutions to Suggested Problems from Section 1.3 1.3.1 Let’s say p is the proportion that my neighbor gets. Then my friend gets 2p, so I get 4p. We know p + 2 p + 4 p = 1, so my part is 4 p = 4 / 7. 1.3.3 Let Ω = { 1 , 2 , . . . , 500 } . Then the sets represented in parts (a), (b), and (c) are A = { 17 , 93 , 202 } , B = Ω \ { 4 , 17 , 93 , 101 , 102 , 202 , 398 } , and C = { 16 , 18 , 92 , 94 , 201 , 203 } . 1.3.4 (a) { 0 , 1 } . (b) { 1 } . (c) Can’t be represented; we only count the number of times we got heads, not the order of the flips. (d) { 1 , 2 } . 1.3.6 Let p ( n ) be the probability that a word picked at random from the given sentence has length n . Then, using the table from problem 1.1.2, we have p (1) = 1 / 10 p (2) = 2 / 10 p (4) = 3 / 10 p (6) = 2 / 10 p (7) = 1 / 10
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Unformatted text preview: / 10 and p ( n ) = 0 for all other numbers. If q ( n ) is the probability that a word picked at random has n vowels, then q (1) = 6 / 10, q (2) = q (3) = 2 / 10, and q ( n ) = 0 for all other numbers. 1.3.9 (a) P ( F ∪ G ) = P ( F ) + P ( G )-P ( FG ) = . 7 + . 6-. 4 (b) P ( F ∪ G ∪ H ) = P ( F ) + P ( G ) + P ( H )-P ( FG )-P ( FH )-P ( GH ) + P ( FGH ), which is . 7 + . 6 + . 5-. 4-. 3-. 2 + . 1 = 1. (c) P ( F c G c H ) = P ( H )-P ( GH )-P ( FH ) + P ( FGH ) = . 5-. 3-. 2 + . 1 = . 1 Last compiled at 2:21 P.M. on October 4, 2008...
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