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Discrete Mathematics Math 6A

Discrete Mathematics Math 6A. Homework 6 Solution. 5.2-2 We are told that p(3)=2p(x) for each x ≠ 3, but it is implied that p(1) = p(2) = p(4) = p(5) = p(6). We also know that the sum of these six numbers must be 1. It follows easily by algebra that p(3)=2/7 and p(x)=1/7 for x = 1,2,4,5,6

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Discrete Mathematics Math 6A

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  1. Discrete MathematicsMath 6A Homework 6 Solution

  2. 5.2-2 We are told that p(3)=2p(x) for each x ≠ 3, but it is implied that p(1) = p(2) = p(4) = p(5) = p(6). We also know that the sum of these six numbers must be 1. It follows easily by algebra that p(3)=2/7 and p(x)=1/7 for x = 1,2,4,5,6 5.2-9 Note that there are 26! permutation of the letters, so the denominator in all of our answers is 26!. To find the numerator, we have to count the number of ways that the given event can happen. Alternatively, in some cases we may be able to exploit symmetry. (a) There are only one way for this to happen, so the answer is 1/26! (b) There are 25! ways to choose the rest of the permutation after the first letter has been specified to be a. Therefore the answer is 25!/26! = 1/26. Alternatively, each of the 26 letters is equally likely to be first, so the prob. that z is first is 1/26 (c) Since z has either to precede a or to follow it, and there is no reason that one of these should be any more likely than the other, we immediately see that the answer is 1/2 (d) In effect we are forming permutation of 25 items – the letters b through y and the double letter combination az. There are 25! ways to do this, so the answer is 25!/26! = 1/26. Here is another way to reason. For a to immediately precede z in the permutation, we must first make sure that z does not occur in the first spot, and the prob. of that is clearly 25/26. Then the prob. that a is the letter immediately preceding z given that z is not first is 1/25, since each of the 25 other letters is equally likely to be in the position in front of z. Therefore the desired prob. is (25/26)/(1/25) = 1/26

  3. (e) We solve this by the first technique used in part (d). In effect we are forming a permutation of 24 items, on of which is the triple letter combination amz. There are 24! ways to do this, so the answer is 24!/26! = 1/650 • (f) If m,n and o are specified to be in their original positions, then ther are only 23 letters to permute, and there are 23! ways to do this. Therefore the prob. is 23!/26! = 1/15600 • 5.2-11 Clearly p(E  F)  p(E) = 0.7. Also, p(E F) 1. If we apply Theorem 2 from section 5.1, we can rewrite this as p(E) + p(F) – p(E  F)  1, or 0.7 + 0.5 – p(E  F)  1. Solving for p(E  F) gives p(E  F)  0.2 • 5.2-24 p(F) = 16/32 and p(EF) = 1/32. Therefore p(E|F) = p(EF)/p(F) = 1/16 • 5.2-27 The event E and F are independent iff p(EF) = p(E)p(F). Let p(E) be the prob. that have children of both sexes and p(F) be the prob. that have at most one boy. • n=2 BB, BG,GB, GG  p(E)= 2/4 and p(F) = ¾ and p(EF)=2/4 p(E)p(F)= 3/8 and p(E)p(F) ≠p(EF). Therefore, two events are not indepent. • n=4 p(E) = 14/16 and p(F)=15/16 and p(EF) = 14/16 and p(E)p(F)=210/256. Therefore, the events are not independent. • n=5  p(E) = 30/32 and p(F) = 31/32 and p(E F)= 30/32 and p(E)p(F) = 930/1024. Therefore, the events are not independent.

  4. 5.2-30 • (1/2)^10 • 0.6^10 • (½)(1/2)^2(1/2)^3...(1/2)^10 = ½^(1+2+...+10) = ½^55 • 5.2-34 b(k;n,p) • k=0 (=all fail)  b(0;n,p) = C(n,0)p0(1-p)n = (1-p)n • at least one success = 1-(all fail) not the case that no successes  1-(1-p)n • at most one success = no successes or one success  • one success = (k=1)  b(1;n,p) = C(n,1)p1(1-p)n-1 = n*p1(1-p)n-1 • Therefore, (1-p)n + n*p1(1-p)n-1, where n>0 • d) at least two failures = 1- (no successes or one success)  1-{(1-p)n + n*p1(1-p)n-1}

  5. 5.3-5 p(3) = 2/7, p(1) = p(2) = p(4) = p(5) = p(6) = 1/7 p(X=2) = 1/7*1/7 = 1/49 : (1,1) p(X=3) = 1/7*1/7 + 1/7*1/7 = 2/49 : (1,2) and (2,1) p(X=4) = 1/7*2/7 + 1/7*1/7 + 2/7*1/7 = 5/49 : (1,3), (2,2) and (3,1) p(X=5) = 1/7*1/7 + 1/7*2/7 + 2/7*1/7 + 1/7*1/7 = 6/49 p(X=6) = 1/7*1/7 + 1/7*1/7 + 2/7*2/7 + 1/7*1/7 + 1/7*1/7 = 8/49 p(X=7) = 1/7*1/7 + 1/7*1/7 + 1/7*2/7 + 2/7*1/7 + 1/7*1/7 + 1/7*1/7 = 8/49 p(X=8) = 1/7*1/7 + 2/7*1/7 + 1/7*1/7 + 1/7*2/7 + 1/7*1/7 = 7/49 p(X=9) = 2/7*2/7 + 1/7*1/7 + 1/7*1/7 + 1/7*2/7 = 6/49 p(X=10) = 1/7*1/7 + 1/7*1/7 + 1/7*1/7 = 3/49 p(X=11) = 1/7*1/7 + 1/7*1/7 = 2/49 p(X=12) = 1/7*1/7 = 1/49 E(X) = 2*1/49 + 3*2/49 + 4*5/49 + 5*6/49 + 6*8/49 + 7*8/49 + 8*7/49 + 9*6/49 + 10*3/49 + 11*2/49 + 12*1/49 = 336/49  6.86

  6. 5.3-6 p(X=$10,000,000) = 1/C(50,6) p(X=$0) = 1- (1/C(50,6)) E(X) = 10,000,000*1/C(50,6) + 0*{1- (1/C(50,6))} = 10,000,000/15,890,700  0.63 5.3-11 Let the random variable X be the number of times we roll the die. For i = 1,2,...,9, the prob. that X = i is (5/6)i-1(1/6), since to roll the die exactly i times requires that we obtain something other than a 6 exactly i-1 times followed by 6 on the i-th roll. Furthermore, p(X=10) = (5/6)9. We need to compute (i=1..10) i*p(X=i) = (i=1..10) i*(5/6)i-1(1/6). A computer algebra system gives the answer is 50700551/10077696  5.03 5.3-12 If X is the number of times we roll the die, then X has geometric distribution with p=1/6 (a) p(X=n) = (1-p)n-1p = (5/6)n-1(1/6) = 5n-1/6n (b) 1/(1/6) = 6 by Theorem 4 5.3-18 Note that by the definition of maximum and the fact that X and Y take on only nonnegative values, Z(s)  X(s) + Y(s) for every outcome s. Then E(Z) = (sS)p(s)Z(s)  (sS)p(s)(X(s) + Y(s)) = (sS)p(s)X(s) + (sS)p(s)Y(s)= E(X) + E(Y)

  7. 5.3-23 In Example 18 we saw that the variance of the number of success in n Bernoulli trials is npq. Here n=10 and p = q = ½. Therefore the variance is 5/2. 5.3-24 In example 18 we was that the variance of the number of successes in n Bernoulli trials is npq. Here n=10 and p=1/6 and q=5/6. Therefore, the variance is 25/18 5.3-25 We assume that the question is asing about the singed difference. For example, if n=6 and we get five tails and on head, then X6=4, where if we get five heads and on tail, then X6 = -4. The key here is to notice that Xn is just n minus twice the number of heads. (a) The expected number of heads is n/2. Therefore the expected value of twice the number of heads is twice this, or n, and the expected value of n minus this is n-n=0. If it were not zero, then there would be a bias favoring heads or favoring tails. (b) Since the expected value is 0, the variance is the expected value of the square of Xn, namely the square of twice the number of heads. This is clearly four times the square of the number of heads, so its expected value is 4*n/4 = n, from example 18, since p=q=1/2. Therefore the answer is 1.

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