select_solutions.pdf

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Department
Mathematics
Course Code
MA340
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Adam Metzler

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3.19(b) Before we actually start the problem let’s get a feel for it in the case n = 5. To avoid confusion assume the boxes are labelled 1;2;3;4;5 and the balls labelled A;B;C;D;E. An outcome can be represented as a 5▯tuple such as (1;1;3;2;1), and we interpret this as A going to box 1, B going to box 1, C going to box 3, D going to box 2 and E going to box 1. There are clearly 5 total outcomes, and it is worth noting that the outcome (1;1;3;2;1) will leave exactly two boxes (boxes 4 and 5) empty. So what is an outcome that leaves exactly one box empty going to look like? Well its going to have to look something like (1;3;1;2;5). We see that two of the balls have to go in the same box and the other three have to go in di▯erent boxes. So let’s break the job of writing down such outcomes into three tasks. First we decide which balls are going to go into the same box, then we decide which box to put them in, then we decide on the boxes for the remaining balls. The ▯rst task amou▯ ▯ to selecting a 2▯element subset of fA;B;C;D;Eg and there are 5 ways to do this, the second task amounts to choosing one of the 2 available 5 boxes to put them in and there are 5 ways to do this. For example we might choose fB;Eg in step one and 4 in step two, leading to a partially completed outcome of ( ;4; ; ;4). To complete this outcome we need only select the boxes that are going to hold the three remaining balls, this amounts to selecting a three-digit permutation of the digits f1;2;3;5g and there are 4 ▯ 3 ▯ 2 = 4! ways to do this. For example we might select the permutation 532, leading to the complete outcome (5;4;3;2;4). ▯5▯ ▯5▯ So there are 2 ▯ 5 ▯ 4! = 2 ▯ 5! total outcomes that leave exactly one box▯em▯ty. In general, ▯i▯h n boxes and n balls, the total number will be n ▯ n ▯ (n ▯ 1)! = n ▯ n! such outcomes. 2 2 3.39(b) Each state has two senators, say A and B. There are 50 states. So we may write the complete list of senators as S = f1 A1 B2 ;A ;:B:;50 ;50Ag : B 1 An outcome here is a 5▯element subset of S such as f4 ;7 A34B;49 A49 g,A B ▯ ▯ and there are 150 total outcomes. In order to write down an outcome that does not repeat a state let us (i) choose the states that will be represented and then (ii) choose which senator (A or B) from each state. Task (i) amounts to selecting a 5▯element subset of f1;2;:::;50g and task (ii) amounts to selecting a 5▯digit string using only the letters A,B (repetitions obviously allowed). For example we might choose f3;9;15;24;42g in step (i) and AABAA in step (ii), leading ▯50 to the committee f3 ;A ;1A ;24B;42 A. ThAre are 5 ways to com- plete task (i) and 2 ways to complete task (ii), therefore there are ▯ ▯ 50 ▯ 2 = 67;800;320 total committees with no state repeated. 5 3.43 We need to ▯ll in the string ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ using 1 P, 4 Es, 2 Rs, 1 S and 1 V . put the letters In order to create such an arrangement we must (i) choose the spot that is going the P, then (ii) choose the four spots that are going to hold the Es, then :::. For example if we choose f4g in step (i) and f1;3;7;8g in step (ii) we get the partially completed outcome E ▯ ▯ E ▯ P ▯ ▯ ▯ E ▯ E ▯ (6) 3.59(b) The answer is 48 ▯4▯45. 52! 3.101 Suppose for the moment that N = 10 and M = 4. Then an outcome will be a string of the form nndnnnddnd, consisting of exactly 4 d’s and 6 n’s. This particular outcome corresponds to the situation where the third, seventh, eighth and tenth items are defective. In order to construct such an outcome we need only perform one task, and that is to select the positions of the 4 d’s. This amounts to selecting a 4-element subset of f1;2;3;:::;9;10g, and the outcome above corresponds to the ▯10 selection f3;7;8;10g, and there are obviously 4 ways to do this. 2 ▯10 So there are 4 outcomes. In general, for arbitrary M and N (with M ▯ N) an outcome will be a string of length N, such as ddnd▯▯▯nndn consisting of exactly M d’s. To construct such an outcome we simply select a M-element subset of f1;2;3;:::;N ▯1;Ng and put the M d’s ▯N▯ in those positions. So there areM total outcomes. th What does an outcome where the k item selected is the last defective look like? Well for N = 10, M = 4 and k = 6 it would have to look something like dndnddnnnn. The general form of such an outcome is ▯ ▯ ▯ ▯ ▯ dnnnn. In order to write down such an outcome we need only choose where, in t▯ ▯▯rst 5 = k ▯ 1 spots, to put the ▯rst 3 = M ▯1 d’s, and there are 5 = k ▯ 1M ▯ 1 ways to do this. So the ▯ 3▯ ▯ ▯ probability of interest isk▯1 = N , assuming of course that k ▯ M M▯1 M (if k < M the probability is zero). ▯M▯▯ N▯M ▯ ▯N▯ 3.109 (a) The probability is k n▯k = n . (b) The probability that we catch a tagged deer is M=N, and we would expect that the proportion of tagged deer in our sample would be reasonably close to this number. Thus we would expect that k=n ▯ M=N and we could estimate N using N ▯ nM=k = 8▯10=3 ▯ 27. So we might estimate that there are approximately 27 deer in the forest. 3 5.40 When sampling without replacement there are (N) n N(N ▯ 1)(N ▯ 2)▯▯▯(N ▯n+1) possible outcomes. In order to determine the number th of outcomes in which the k person selected has the speci▯ed attribute note that there are M ways to select this person, and (N ▯1) ways n▯1 to select the remaining people. Therefore the probability that the k person is selected has the speci▯ed attribute is M ▯ (N ▯ 1) M ▯ (N ▯ 1) ▯ (N ▯ 2)▯▯▯((N ▯ 1) ▯ (n ▯ 1) + 1) n▯1 = (N) n N ▯ (N ▯ 1)▯▯▯(N ▯ n + 1) M ▯ (N ▯ 1) ▯ (N ▯ 2)▯▯▯(N ▯ n + 1) = N ▯ (N ▯ 1)▯▯▯(N ▯ n + 1) M = : N If the sampling is carried out with replacement the total number of n th outcomes is N . As before there are M ways to choose the k person but there are now Nn▯1ways to choose the remaining people. Therefore the probability the khperson that is selected is n▯1 M ▯ N = M : N n N 5.68 Think of the words the students knows as \Type 1." The probability in part (a) is then ▯ ▯▯ ▯ 30;00050;000 4▯ ▯0 = 0:0711937::: : 80;000 100 The probability in part (b) is 40 ▯30;00050;000 X k 100▯k ▯80;000 = 0:4645::: k=35 100 According to the book’s rule of sum the binomial approximation would be appropriate since the student knows more than 5% of the words in the dictionary. Note that if we use the binomial approximation in part 4 (a) we get ▯ ▯▯ ▯ 30;00050;000 ▯ 100▯▯ ▯3▯ 40 5 60 4▯ ▯60 ▯ = 0:0711599::: ; 80;000 40 8 8 100 which is quite a good agreement. For (d) I would estimate the size of the students vocabulary at 38 100 ▯ 80;000 = 30;400 : 5.116 The pmf is ▯ k ▯ 1▯ f(k) = (:67) (:33)▯3 ; k = 3;4;5;::: 2 5 6.6 (a) Start with a probability such P(X = 3;Y = 5). There are two ways this can occur - we get a three on the ▯rst roll and a ▯ve on the second roll, or we get a ▯ve on the ▯rst roll and a three on the second roll. Each of these outcomes has probability 1 , so 36 P(X = 3;Y = 5) = 1 . 18 Now consider a probability such as P(X = 3;Y = 3). The only way this can happen is if we get a three on both rolls, so P(X = 3;Y = 3) = 1. 36 Finally, consider a probability such as P(X = 3;Y = 2). By de▯nition the maximum exceeds the minimum, and so there is now way the minimum can be three when the maximum is two. Therefore P(X = 3;Y = 2) = 0. To summarize, the joint probability mass function of (X;Y ) is 8 1 < 36 if 1 ▯ x = y ▯ 6 ; 1 fX;Y (x;y) = P(X = x;Y = y) = 18 if 1 ▯ x < y ▯ 6 ; : 0 otherwise (b) In tabular form we have X 1 2 3 4 5 6 f Y 1 1 0 0 0 0 0 1 36 36 2 1 1 0 0 0 0 3 18 36 36 Y 3 1 1 1 0 0 0 5 18 18 36 36 4 1 1 1 1 0 0 7 18 18 18 36 36 5 1 1 1 1 1 0 9 18 18 18 18 36 36 6 1 1 1 1 1 1 11 18 18 18 18 18 36 36 fX 11 9 7 5 3 1 36 36 36 36 36 36 (c)-(h) The general idea for problems like this is to identify all points in the range that satisfy the given condition. The upper left panel of Figure 1 illustrates the range of (X;Y ), i.e. the set of all points 6 (x;y) for which f (x;y) > 0, while the other three panels il- X;Y lustrate (in red) the points in the range that satisfy the given condition for parts (f), (g) (the others are left to you). Once we’ve identi▯ed the relevant points in the range we simply sum the values of the joint pmf evaluated at those points. For example 1 P(X = Y ) = f X;Y (1;1)+f X;Y (2;2)+f X;Y (3;3)+f X;Y (4;4)+f X;Y (5;5)+f X;Y (6;6) = ; 6 and 1 P(Y = 2X) = f (1;2) + f (2;4) + f (3;6) = ; X;Y X;Y X;Y 6 etc. Figure 1: Range of (X,Y) Points in Range where X=2X in Red 7 7 6 6 5 5 4 4 y y 3 3 2 2 1 1 00 2 4 6 00 2 4 6 x x 7 9.26 The desired probability is Z 3=4Z 1 Z 3=4▯Z 1 ▯ (x + y)dy dx = (x + y)dy dx 1=4 1=2 1=4 1=2 Z 3=4▯ 2▯y=1 y = xy + 2 dx 1=4 y=1=2 Z 3=4▯ ▯ = x + 3 dx 2 8 1=4 7 = : 16 9.172 The domain of the pair (X;Y ) is the rectangle [0;1]▯[0;2]. The region where the total loss exceeds one is the portion of this rectangle that lies above the line y = 1 ▯ x (why?). Thus the desired probability is Z 1Z 2 (2x + 2 ▯ y)=4dy dx : 0 1▯x 9.178 (c) gives the desired probability. This is because, in order that the device function, both variables must exceed 0.5. 9.190 (a) The marginal pdf of X is Z x f (x) = 8xy dy = 4x ; 0 < x < 1 : X 0 (b) The marginal pdf of Y is Z 1 2 fY(y) = 8xy dx = 4y(1 ▯ y ) ; 0 < y < 1 : y 6.48 Using the table from the solution to 6.6 we get the following table. The ▯rst row corresponds to the conditional mass function f (x;1) = XjY fX;Y(x;1)=fY(1) and gives the conditional distribution P(X = xjY = 1). The second rwo contians f XjY(x;2) = f X;Y(x;2)=fY(2) and gives 8 the conditional distribution P(X = xjY = 2), etc. X 1 2 3 4 5 6 1 1 0 0 0 0 0 2 2 1 0 0 0 0 2 2 1 Y 3 5 5 5 0 0 0 2 2 2 1 4 7 7 7 7 0 0 2 2 2 2 1 5 9 9 9 9 9 0 2 2 2 2 2 1 6 11 11 11 11 11 11 A similar tanble can be constructed for conditional distributions of Y , given X, in which case each column would correspond to a conditional pmf. 9.64 In each case the probability of interest is Z 1 fXjY (x;0:8)dx : 0:9 (a) Here we have f (y) = 1 for 0 < y < 1 (i.e. the marginal distribu- Y tion of Y is uniform) and so f (x;0:8) = 1 (i.e. the conditional XjY distribution of X, given Y = 0:8, is uniform) the probability of interest is 0:1. R 1 (b) Here the marignal density of Y is f (y)Y= 0 (x + y)dx = 0:5 + y x+0:8 and so f XjY (x;0:8) = [x + 0:8]=[0:5 + 0:8] = 1:3. The proba- 1▯(0:9) +(0:8)(0:1) bility of interest is therefore 1:3 = 0:207:::. This is larger than what we got in (a) because the conditional distribu- tion of X here is more heavily weighted towards larges values of X. 1 3 3 (c) Here the marginal density of Y is f (y)Y= 2 + 2 , leading to the conditional density 3x + 3(0:8) 2 fXjY (x;0:8) = ; 1 + 3(0:8) 2 9 and the probability of interest is 1 ▯ (0:9) + 3(0:8) (0:1) 2 = 0:158::: 1 + 3(0:8) 9.69 Ignore part (a), hard to see why there is a di▯erence until we work through the example. You are encouraged to sketch T, noting that it is a triangle bounded below by the horizontal axis, above by the line y = x and on the right by the vertical line at x = 1. (b) In both cases the joint density is positive for all (x;y) 2 T, i.e. all (x;y) such that 0 < y < x < 1. For Method I the joint density is simply 1 1 fX;Y(x;y) = = = 2: Area(T) 1=2 For Method II the joint density is obtained as follows 1 1 fX;Y(x;y) = fY jX(y;x) ▯Xf (x) = ▯ 1 = : x x (c) For Method I we have Z x fX(x) = 2dy = 2x ; Z0 1 fY(y) = 2dx = 2(1 ▯ y) ; y and for Method II we have Z x f (x) = 1 dy = 1 ; X 0 x Z 1 1 fY(y) = x dx = ▯ln(y) : y (d) For Method I the relevant conditional pdfs are 2 1 fXjY(x;y) = 2(1 ▯ y) = 1 ▯ y ; y < x < 1 ; 10 so that the conditional distribution of X, given Y = y, is uniform on (y;1), and f (y;x) = 2 = 1 ; 0 < y < x ; Y jX 2x x so that the conditional distribution of Y , given X = x, is uniform on (0;x). For Method II we have 1=x 1 fXjY (x;y) = ▯ln(y) = ▯ xln(y) ; y < x < 1 ; and f (y;x) = 1=x = 1 ; 0 < y < x ; Y jX 1 x so that the conditional distribution of Y , given X = x, is uniform on (0;x). In both cases the conditional distribution of Y , given X, is uni-
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