Math , Fall 2012: HW 1 Solutions


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1 Math 3., Fall : HW Solutions Problem (p.9 #). Suppose a wor is picke at ranom from this sentence. Fin: a) the chance the wor has at least letters; SOLUTION: All wors are equally likely to be chosen. The sentence has wors; 7 are at least letters long. We have P (at least letters) = 7. b) the chance that the wor contains at least vowels (a,e,i,o,u); SOLUTION: Assuming that the vowels nee not be istinct (i.e. the three instances of e in sentence count as three vowels in the wor), then exactly wors contain at least two vowels, so P (at least vowels) =. c) the chance that the wor contains at least letters an at least vowels. SOLUTION: Exactly four wors contain at least letters an at least two vowels, so P (at least letters an at least vowels) =. ) What is the istribution of the length of the wor picke? SOLUTION: Let i be the length of the wor. Then the values of i which occur with nonzero probability are i =,,, 6, 7, an 8. We have P () =, P () =, P () = 3, P (6) =, P (7) =, P (8) =. e) What is the istribution of the number of vowels in the wor? SOLUTION: Let j be the number of vowels in the wor. Then the values of j which occur with nonzero probability are j =,, 3. We have P () = 6, P () =, P (3) =. Problem (p.9 #7). Suppose two sie ice are rolle. Fin the probabilities of the following events: a) the maximum of the two numbers rolle is less than or equal to ; SOLUTION: The sample space is the set of orere pairs (i, j), where i an j are integers between an, an i represents the number on the upper face of the first ie an j the number on the upper face of the secon ie. All outcomes are equally likely. There are 6 outcomes. The outcomes (, ), (, ), (, ), an (, ) are the outcomes whose maximum is less than or equal to. Hence P (maximum is less than or equal to ) =. b) the maximum of the two numbers rolle is less than or equal to 3; SOLUTION: In orer for the maximum of the two numbers to be less than or equal to 3,
2 both numbers must be less than or equal to 3. There are 3 3 = 9 ways for this to happen, so P (maximum is less than or equal to 3) = 9 6. c) the maximum of the two numbers rolle is equal to 3; SOLUTION: In orer for the maximum to be equal to 3, at least one number must be 3, an the another number can be anything less than or equal to 3. There are two choices for which ie results in a 3; once this is fixe, there are choices for the outcome of secon ie to be strictly less than 3, yieling (3 ) outcomes in which the one number is 3 an the other is strictly less than 3. Finally there is the outcome in which both numbers are 3. So there are 3 outcomes in which the maximum is exactly 3, so the probability of this event is 5 6. ) Repeat part c for the maximum equal to,, an. SOLUTION: The solution to this part is ientical to the solutions above; P (max = ) = 6, P (max = ) = 3 6, P (max = ) = 7 6. e) If M is the maximum of the two numbers, then P (M = ) + P (M = ) + P (M = 3) + P (M = ) =, check that your answers for c) an ) satisfy this relationship. SOLUTION: Inee, if the maximum is equal to i, then the upper face of at least one ie must be i (there are two possibilities for which ie will have i on its upper face) an the other ie must result in a number strictly less than i or exactly equal to i. Now, there are (i) numbers strictly less than i, an number exactly equal to i, so we get (i ) + i = (i) possibilities for a maximum exactly equal to i, where i ranges from to. Hence P (i) = i 6, an evaluating this at each of i =,, 3 an an then summing gives the esire result. Problem 3. Now suppose that the ie has n sies: a) the maximum of the two numbers rolle is less than or equal to ; SOLUTION: Again, there are n possible outcomes, all equally likely, an outcomes in which the maximum of the two numbers is less than, so the esire probability is n. b) the maximum of the two numbers rolle is less than or equal to i {,... n}; SOLUTION: As in the previous problem, for the maximum to be less than or equal to i, both ie rolls must result in numbers less than or equal to i. There are i choices for the outcome of the first ie an i choices for the secon, so i ways in total for the maximum of the two numbers to be less than i. The esire probability is i n. c) the maximum of the two numbers rolle is equal to i {,..., n};
3 SOLUTION: As in the previous problem, one of the numbers must be equal to i an the other must be less than or equal to i. If the first one is equal to i, the secon number can take on i values; if the secon is equal to i, the first can take on i values, an the orere pair (i, i) shoul not be counte twice. Hence there are i ways for the maximum to be equal to i. The esire probability is therefore i n. Problem. Consier that the ata {.,.,.,.,.75,.,.5,.6,.9,.99, 3.5, 3.5, 3.3, 3.7} an the cut points {,,, 3, }. Draw the empirical istribution for this ata. Label the height for each rectangle as in Section.3. What is the total area of all of the rectangles? SOLUTION. We note that the ifference between each consecutive pair of cut points (b j, b j+ ) is, an there are ata points in total. For each pair (b j, b j+ ) of cut points, where the height of the rectangle whose base is the interval between (b j, b j+ ) is given by H j = # of ata points x i for which b j < x i < b j+. [total number of ata points][b j+ b j ] So, for instance, the height above the interval (, 3) is 5. The total area of all the rectangles sums to, because the height of each rectangle is the fraction of ata points that lie in the given interval an the with of each rectangle is. Problem 5 (p.3 #). Events A, B, an C are efine on an outcome space. Fin expressions for the following probabilities in terms of P (A), P (B), P (C), P (AB), P (AC), P (BC), an P (ABC). a) The probability that exactly two of the A, B, C occur. SOLUTION: We want to fin the probability of the events E = A B C c = ABC c, E = A C B c, an E 3 = B C A c. Each of these events E i represents exactly two of A, B, or C occurring. Note that these events have no intersection. Therefore, P ({A B C c } {A C B c } {B C A c }) = P (E ) + P (E ) + P (E 3 ). Now, P (E ) = P (AB) P (ABC), P (E ) = P (AC) P (ABC), an P (E 3 ) = P (BC) P (ABC). Hence we get P (exactly two of the events occur) = P (AB) + P (AC) + P (BC) 3P (ABC). b) The probability that exactly one of these events occurs. SOLUTION: Here we want to fin the probabilities of the events F = AB c C c, F = BA c C c, an F 3 = CA c B c. Again, the events F i are mutually exclusive an their union is event that exactly one of the events occurs. We fin P (F ) = P (A) P (AB) P (AC) + P (ABC); P (F ) = P (B) P (AB) P (BC) + P (ABC); P (F 3 ) = P (C) P (AC) P (BC) + P (ABC) which yiels P (A) + P (B) + (C) P (AB) P (BC) (AC) + 3P (ABC) as the answer. 3
4 c) The probability that none of these events occur. SOLUTION: The probability that no event occurs is the complement of the probability that at least one event occurs. The probability that at least one event occurs is P (A B C). By the inclusionexclusion formula, we have P (A B C) = P (A) + P (B) + P (C) P (AB) P (AC) P (BC) + P (ABC) so the probability that none of the events occur is P (A B C), which by inclusionexclusion is [P (A) + P (B) + P (C) P (AB) P (AC) P (BC) + P (ABC)]. Problem 6. How many chilren shoul a family plan to have so that the probability of having at least one chil of each sex is at least.95? (Assume that both sexes are equally likely.) SOLUTION: Let n be the number of chilren that the family will have, an let A n be the event that there is at least one boy an one girl among the n chilren. It is easier to compute the probability of the event, A c n, that all n chilren are of the same sex. We want to fin n so that: P (A n ) = P (A c n).95 P (A c n).5 Observe that the events M n = n boys, an F n = n girls are a partition of A c n, an P (M n ) = P (F n ) = / n. So we solve: P (A c n) = n.5 n n log The smallest value which satisfies this inequality is n = 6. Problem 7. Suppose we roll two fair sixsie ice. What is the istribution of the ifference (in absolute value) between the two numbers on the top faces? SOLUTION: The smallest possible ifference is an the largest is 6 = 5. When the ifference is, the numbers must be equal there are 6 ways for this to occur. When the ifference is, there are 5 possibilities for the smaller number (6 cannot be the smaller of the two numbers, an the larger number is etermine by the smaller by aing ), an for each possibility, there are two outcomes (e.g. (, ) an (, )), giving ways for the ifference to be. By similar reasoning, there are = 8 outcomes with a ifference of, 3 = 6 outcomes with a ifference of 3, an so on. The istribution is then ifference, i 3 5 P (i) 6 36 = The next problem is a classic puzzle.
5 Problem 8. Three people are going to play a cooperative game. They are allowe to strategize before the game begins, but once it starts they cannot communicate with one another. The game goes as follows. A fair coin is tosse for each player to etermine whether that player will receive a re hat or a blue hat, but the color of the hat (an result of the coin toss) is not reveale to the player. Then the three players are allowe to see one another, so each player sees the other two players hats, but not her own. Simultaneously, each player must either guess at the color of her own hat or pass. They win if noboy guesses incorrectly an at least one person guesses correctly (so they can t all pass). The players woul like to maximize their probability of winning, so the question is what shoul their strategy be? A naive strategy is for them to agree in avance that two people will pass an one person (esignate in avance) will guess either re or blue. This strategy gives them a 5% chance of winning, but it is not optimal. Devise a strategy that gives the players a greater probability of winning. SOLUTION: First, write out the outcome space for the assignment of hats. Ω = {RRR, RRB, RBR, BRR, RBB, BRB, BBR, BBB} The goal is not to win for every outcome, but to win for more than half of the outcomes. Notice that in all but two of the outcomes, two people have the same color hat an one person has a ifferent colore hat. In each of these scenarios, then, one person sees two likecolore hats, while the other two people see one hat of each color. For example, if the outcome is RRB, then the person with the blue hat sees two re hats on the other players heas, an each person with a re hat sees a re an a blue hat. Therefore, the following strategy will guarantee a win whenever two people have the same colore hat an one person has the other color: If a player sees two hats of the same color, then guess the opposite color; If a player sees two ifferent colore hats, then pass. This strategy always fails when all three hats are the same color, since all three players will guess incorrectly. Therefore, the probability of winning with this strategy is 6/8 = 75%. Now for a little review of calculus. These are things you shoul have learne in a class before this one. If you have problems with them, get help now to refresh your memory. Problem 9. Do the following integrals: x 3 x x exp(x)x exp( x /)x SOLUTION. For the first integral, by the funamental theorem of calculus an the power rule, x 3 x = x = For the secon integral, note that the integran is the prouct of two functions: x an exp(x), one of which gets simpler with ifferentiation an the other of which is easy to integrate. Hence this type of integral can be aresse through integrationbyparts. Also, as x, x exp(x) (write the expression as a quotient an use L hôpital s rule). We erive x exp(x)x = [x exp(x)] exp(x)x = 5
6 The last integral requires a bit of ingenuity. Let us first observe that exp( x /)x = exp( y /)y; that is, in a efinite integral, the particular choice of letter to represent the variable of integration is irrelevant. Hence we get [ ] [ ] [ ] exp( x /)x Now, consier the ouble integral exp( y /)y = exp [ (x + y )/ ] xy exp( x /)x This is an iterate integral first we integrate with respect to the variable x, an then with respect the variable y. A priori, we cannot say that the iterate integral an the prouct of the two separate integrals are equal. But since exp [ (x + y )/ ] = exp( x /) exp( y /), the function f(x, y) = exp [ (x + y )/ ] can be ecompose into a prouct of functions f(x, y) = g(x)h(y), where g(x) = exp( x /), an h(y) = exp( y /). While integrating f(x, y) = g(x)h(y) first in the variable x, we can treat h(y) as a constant. Namely, we have exp [ (x + y )/ ] [ ] xy = exp( y /) exp( x /)x y [ ] [ ] = exp( x /)x exp( y /)y = [ ] exp( x /)x From the above chain of equalities, we see that if we can evaluate the ouble integral over the whole plane, then we can etermine the value of the integral we want. The ouble integral over the whole plane exp [ (x + y )/ ] xy can be evaluate by changing to polar coorinates: x = r cos θ, y = r sin θ, where r <, θ π. By the changeofvariables formula for multiple integrals, in a change of variables from (x, y) to (r, θ), the area ifferential xy is transforme to the ifferential rθ multiplie by the absolute value of the eterminant of the Jacobian matrix of the changeofcoorinate map (x,y) (r,θ). You can compute this matrix (in fact you shoul!), an you will fin its eterminant to be r, which is always nonnegative. Hence the area ifferential xy is transforme uner this change of variable to rrθ. The ouble integral in the new (r, θ) coorinates is exp [ (x + y )/ ] xy = π Since the function being integrate oes not epen on θ, this becomes π exp [ r / ] rr exp [ r / ] rrθ 6
7 which can be evaluate by simple substitution. The final answer for the ouble integral is π, so the final answer for the original integral is π. Problem. Perform the following ifferentiations: x (x ) x (x exp( x)) x (ln(x )) SOLUTION. These are straightforwar an we omit the etails: the first ifferentiation is an application of the power rule; the secon an application of the prouct rule; the thir an application of the chain rule. We get x (x ) = x 3 x (x exp( x)) = (x x ) exp( x) x (ln(x )) = x Problem. Evaluate the following infinite sums: ( ) k 3 k k! k= SOLUTION. The first series is the sum of a geometric series whose first term is an whose ratio is. Hence its sum is / / = / 3/ = /3. The secon series is almost the series expansion for the exponential function exp(x), except the first term in that series, 3! =, is not inclue. Hence the numerical value of the infinite sum is exp(3). Problem. Draw a picture of the region of the xyplane were both x an y are between an an y x. Fin the area of this region. SOLUTION. The area is represente by the integral Problem 3. Evaluate the integral k= [ x ]x = [x x 3 /3] = /3 y e x xy Hint: Draw the region over which you are integrating an change the orer of integration. SOLUTION. The region is {y/ x, y }. In orer to change the orer of integration, we nee to express the limits first in terms of y (so the y limits epen on x), an then in terms of x, so the x limits are then two numbers. We see that if y/ x an y, then y x, an x. (Draw these two regions separately an check that they are the same region!) By interchanging the orer of integration, we get y e x xy = = x xe x x e x yx = exp(x ) = exp() 7
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