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Berkeley MATH 55 - Math 55 Solutions to Problems

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Math 55 Solutions to Problems in H W Lenstra s Notes July 9 1999 7 38 am Here are solutions to all 27 problems at the end of the notes on Probability theory by H W n k Lenstra Jr 1988 His combinatorial coefficient 1 P Andrew P Beatrix P Charles 0 1 2 1 2 1 3 1 3 0 0 1 2 1 3 1 6 1 1 2 0 1 3 1 2 is here rendered nCk n k n k 2 Without a Joker the independent pairs are A B and A C but not B C With a Joker in the deck no pair is independent 3 Let P A and P B so that P A 1 and P B 1 Since A and B are independent P A B Then P A P A B P A B P A B whence follows P A B 1 Similarly P A B 1 and P A B 1 1 Thus A and B are independent because P A B P A P B Likewise for A and B But A and A cannot be independent since P A A P 0 unless 0 or 1 4 Yes independent because P divisible by 3 P divisible by 5 1 3 1 5 1 15 P divisible by 3 and by 5 Then P GCD number 15 1 1 1 3 1 5 1 15 8 15 5 5000 700 3 100 10000 0 60 6 For any number n the probability that n balls of any particular color will be drawn is the same for every color because of the situation s symmetry permuting the color s names does not change their probabilities Therefore the Expected number of balls drawn is the same for every color Since all three colors Expected numbers sum to 12 each Expected number is 4 7 a f x x mod 3 and g x x mod 4 so x 4 f x 9 g x mod 12 by the Chinese Remainder Theorem Since 900 0 mod 12 as s ranges over the set S 1 2 900 with uniform probability 1 900 per element P f s m 1 3 for each m in 0 1 2 and P g s n 1 4 for each n in 0 1 2 3 and then P f s m and g s n P s 4 m 9 n mod 12 1 12 because 4 m 9 n mod 12 runs through all 12 members of 0 1 2 11 as m n runs through all 12 pairs Therefore f and g are independent 7 b E f g E f E g 1 3 2 5 2 E f g E f E g 1 3 2 3 2 and Variance V f g V f V g 2 3 5 4 23 12 because f and g are independent 8 E emales 9 3 3 V emales 9 2 9 2 assuming independence of sex of each birth 9 The number of ways of choosing 5 volumes out of 10 is 10 C5 252 The number of ways to get no complete novel is 25 32 The number of ways to get one complete novel is 5C 4C 23 160 The number of ways to get two complete novels is 5C 3C 21 60 As a 1 3 2 1 check we observe that 60 160 32 252 Assuming each way as likely as every other P i 0 32 252 8 63 P i 1 160 252 40 63 P i 2 60 252 15 63 E i 70 63 10 9 Variance i 200 567 Prof W Kahan Page 1 This document was created with FrameMaker 4 0 4 Math 55 Solutions to Problems in H W Lenstra s Notes July 9 1999 7 38 am 10 This solution takes advantage of three identities obtained by differentiating the first twice 1 1 q n 0 qn 1 1 q 2 n 0 n qn 1 2 1 q 3 n 0 n n 1 qn 2 V f n 0 qn 1 p n 1 p 2 n 0 qn 1 p n2 2 n 0 qn 1 n n 0 qn 1 p p q n 0 n n 1 qn 2 p 2 n 0 n qn 1 m 0 qm p recall 1 q p 2 p q p3 p 2 p2 1 p2 2 q p 2 1 p2 q p2 11 Any positive integer n and nonnegative fraction p 1 determine a Binomial Random Variable f it is the count of the successes in n independent Bernoulli trials each with probability p of success P f k nCk pk 1 p n k Now P f is even 0 j n 2 nC2j p2j 1 p n 2j and P f is odd 1 P f is even In the special case that p 1 2 these expressions simplify to P f is even P f is odd 1 2 as is obvious if n is odd because then each term nC2j 2 n included in the sum can be paired with an equal term nCn 2j 2 n excluded from that sum if n is even the simplification of the sum is not obvious However the 2n equally likely outcomes of n trials can be put into pairs that differ only in their first trials since the members of each pair have opposite even odd parity the odd counts f must be as numerous as the even counts and equally likely 12 The same reasoning as worked in Example 7 implies that 2 is the expected number of children whose first name starts with the letter that they get The variance cannot be determined because it is positive unless all children s names begin with the same letter in which case the variance is zero 13 There are 2100 1 26765 1030 subsets of a set with 100 elements The subsets with cardinalities between 41 and 59 inclusive number 41 k 59 100Ck 100C50 2 1 k 9 100C50 k 1 19554 1030 Their ratio is 1 19554 1 26765 0 9431 This ratio is the same as the probability that a binomial random variable counting the successes in n 100 independent Bernoulli trials with p 1 2 will depart from the mean n p 50 by less than 10 5 2 times the standard deviation n p 1 p 5 Chebyshev s Inequality says this probability is at least 1 1 22 0 75 The Central Limit Theorem uses a Normal random variable see the class notes titled Law of Large Numbers distributed continuously with the same mean and standard deviation to estimate that a departure from the mean smaller than times the standard deviation has probability near but what value of should be used for the given discrete distribution If 10 5 2 then 0 955 overestimates the probability If 9 5 1 8 then 0 928 underestimates the probability If 9 5 5 1 9 then 0 943 gets it about right Prof W Kahan Page 2 …


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