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hw07sol

Course: CS 280, Fall 2003
School: Cornell
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Word Count: 770

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280 CS Solution Set Homework 7 by Shaddin Doghmi 1(a). 300 Bernoulli Trials. Probability of success p=1/6. We know that E (# of successes)=E(Y6)=np (see book p.277) E(Y6) = np = 3001/6 = 50 1(b). V(Y6)=npq (see book p.282) =3001/65/6=125/3 1(c). 1(d). E(Y1 + Y2+ Y3) = E(Y1) + E(Y2) + E(Y3)= 50 + 50+ 50 =150 Make random variables X1, X2 ..... X300 . Xi=1 if trial i yields 1,2,3 . Xi=0 otherwise. Y1+Y2+Y3 = X1...

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280 CS Solution Set Homework 7 by Shaddin Doghmi 1(a). 300 Bernoulli Trials. Probability of success p=1/6. We know that E (# of successes)=E(Y6)=np (see book p.277) E(Y6) = np = 3001/6 = 50 1(b). V(Y6)=npq (see book p.282) =3001/65/6=125/3 1(c). 1(d). E(Y1 + Y2+ Y3) = E(Y1) + E(Y2) + E(Y3)= 50 + 50+ 50 =150 Make random variables X1, X2 ..... X300 . Xi=1 if trial i yields 1,2,3 . Xi=0 otherwise. Y1+Y2+Y3 = X1 + X2 ..... X300 Because Xi are all pairwise independent(different trials): V(Y1+Y2+Y3)= V(X1) + V(X2) +....... +V(X300) V(Xi)=E(Xi2)-(E(X)) 2=(1/212 + 1/202)-(1/21+ 1/2 0) 2=1/2-1/4=1/4 V(Y1+Y2+Y3) = 3001/4 = 75. 2(a). E(Y1)= E(Y2).....=E(Y6) E(X)= E(Y1)+2E(Y2) ...... + 6E(Y6)=2150 = 1050 Make random variables X1, X2 ..... X300 . Xi = k if trial i yields k. X=X1+ X2 ...... +X300 Because Xi are all pairwise independent (depend on different trials) V(X)= V(X1) +V(X2)+.....+V(X300) V(Xi)=1/6 (1+4+9....36) (1/6 (1+2+3...+6))2= 35/12 V(X)=300 V(Xi) = 875 2(b). 3. For each fruit, we use bars and stars to divide the fruit amongst 4 people. Therefore we get: Number of ways to split up the fruits= The number of ways to split up apples number of ways to split up bananas number of ways to split up pears= 7!/(3! 4!) * 6!/(3! 3!) * 5!/(3! 2!)= 7000 4(a). Two ways to do this: Solution1: Take Computer science books to be |, and non Computer Science books to be *. Since there are 5 bars, there are 6 bins in which we can place the 14 *'s. The middle 4 bins (between pairs of |'s) must have at least one * each in them. So, after placing 4 *'s in the middle 4 bins, we now have 10 *'s left to place in the 6 bins. The number of ways to place 10 stars in 6 bins is C(15,5). Each arrangement of *'s and |'s is a choice of CS book positions. There are C(15,5) ways to choose the CS book positions. Now, for each choice of positions, lets permute the CS books by multiplying by 5!, and lets permute the non CS books by multiplying by 14!. So our answer is: C(15,5) 5! 14! = (15! 14!) / 10! Solution 2: Take Computer science books to be |, and non Computer Science books to be *. The number of ways we can arrange those is the number of ways we can arrange the bars and stars such that there is at least one star between each two bars. One way to do this would be to make 4 Bar-Star pairs ( |*), lets denote those by _. We then have 10 *'s, 4 _'s, and 1 |. We know the 4 _'s have come to before the |, so choosing the position of the CS books involves C(15,5). This is the number of ways we can choose 5 positions for 4 _'s and a |, in that order. There are C(15,5) ways to choose the CS book positions. Now, for each choice of positions, lets permute the CS books by multiplying by 5!, and lets permute the non CS books by multiplying by 14!. So our answer is: C(15,5) 5! 14! = (15! 14!) / 10! 4(b). divide by the number of permutations of each indistinguishable group. We get: (15! 14!) / (10! 8! 6! 5!) 5. Inclusion-Exclusion: Ak= Integers less than 4445 divisible by k |Ak|= 4444/k A= integers divisible by 2,3,5 or 7 |A|= |A2| + |A3|+|A5| + |A7| -|A2 A3| - | A2 A5| - |A2 A7| - | A3 A5|- |A3 A7| - | A5 A7| + |A2 A3 A5| + |A2 A3 A7| + |A2 A5 A7|+ |A3 A5 A7| - |A2 A3 A5 A7| = |A2| + |A3|+|A5| + |A7| -|A6| - | A10| - |A14| - | A15|- |A21| - | A35| + |A30| + |A42| + |A70|+ |A105|-|A210| = 2222 + 1481+ 888+634 740 444 317 296 211 126 + 148 + 105 + 63 + 42 21 = 3428 6(a) Permutations of the beads (taking into account that yellow are identical)= 23!/20! Divide by 23 to account for rotation (23 rotations) Also divide by 2 to account for mirror image of the permutation(turn over). # of different necklaces= 23!/(20! 23 2)=((22 21)/2) = 231 6(b) Permutations of the beads (taking into account that yellow are identical and black are identical)= 23!/(20! 3!) Divide by...

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