# m (t) = e nt m Y ( t) = e nt (pe t + q) n = (pe t e t + qe t ) n = (qe t + p) n

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1 1. For a discrete random variable Y, prove that E[aY + b] = ae[y] + b and V(aY + b) = a 2 V(Y). Solution: E[aY + b] = E[aY] + E[b] = ae[y] + b where each step follows from a theorem on expected value from Chapter 3. V(aY + b) = E [((ay + b) E[aY + b]) 2 ] = E [(ay + b ae[y] b) 2 ] = E [a 2 (Y E[Y]) 2 ] = a 2 E [(Y E[Y]) 2 ] = a 2 V(Y) 2. (a) If Y is a random variable with moment generating function m(t), show that for a random variable W = ay + b, its moment generating function is m W (t) = e bt m(at). Solution: m W (t) = E[e t(ay+b) ] = E[e bt e (at)y ] = e bt E[e (at)y ] = e bt m Y (at) (b) Using the result in part (a), show the following: If Y is a binomial random variable with parameters n and p, and Y = n Y, then the moment generating function of Y is m (t) = (qe t + p) n where q = 1 p. Based on that moment generating function, what distribution does Y have? Explain why this makes sense. Solution: For a binomial random variable Y, m Y (t) = (pe t + q) n. Using part (a) with a = 1 and b = n gives m (t) = e nt m Y ( t) = e nt (pe t + q) n = (pe t e t + qe t ) n = (qe t + p) n which is the moment generating function for a binomial random variable with parameters q and n. This makes sense because Y is the number of failures in n trials where failure occurs with probability q. 3. A five card poker hand is being dealt one card at a time from a deck of 52 cards. If the first two cards dealt form a pair, what is the probability that the player eventually gets four-of-a-kind in their hand?

2 Solution: The number of ways to get three more cards from the fifty cards remaining in the deck is ( 5 ). 3 There are only 48 ways to get four-of-a-kind since you just have to pick which of the 48 other cards will fill out your hand. So, the probability of eventually getting four of a kind is 48 ( 5 = 48 3 ) 196 = A weatherman is 9% accurate at predicting rain. That means that on days when it does rain, there is a 9% chance that the weatherman predicted rain on that day and on days when it doesn t rain, there is a ten percent chance that he predicted rain on that day. (a) In Waverly where it rains approximately 9 days per year, if the weatherman predicts rain, what is the probability that it will actually rain? Solution: Consider the events R: it rains, R: it doesn t rain and W: rain is predicted. The information in the problem is that P(W R) =.9 and P(W R) =.1. For Waverly, = 9/365. We are looking for P(R W) and will use Bayes Rule P(W R) P(R W) = P(W R) + P(W R)P( R).9(9/365) =.9(9/365) +.1(275/365) = (b) In San Diego, CA where it rains only 2 days per year, if the weatherman predicts rain, what is the probability that it will actually rain? Solution: As above, but using = 2/365 P(R W) =.9(2/365).9(2/365) +.1(345/365) = = (c) Compare your answers to parts (a) and (b). Solution: Although the accuracy is the same, due to the lack of rainy days in San Diego, the weatherman s predictions are much less useful than in Waverly since in San Diego, he is wrong 66% of the time!

3 5. Suppose that you have three programmers writing computer code for a project: Alice has designed 6% of the code, Barb 3% and Chuck 1%. Suppose further that Alice has a bug in 3% of her work, Barb in 7% of her work, and Chuck in 5% of his. What is the probability that the project contains a bug? Given that you found a bug, who is most likely to have been responsible? Solution: For a given line of code, consider the events A: Alice wrote it, B: Barb wrote it and C: Chuck wrote it. Then we know that P(A) =.6, P(B) =.3 and P(C) =.1. If R is the event that the line has a bug in it then P(R A) =.3, P(R B) =.7 and P(R C) =.5. Since A, B and C form a partition of the entire project, we can use the law of total probability = P(R A)P(A) + P(R B)P(B) + P(R C)P(C) =.6(.3) +.3(.7) +.1(.5) =.44 Given that R occurred, we can compute the probability that each person was responsible P(A R) P(A R) = P(B R) P(B R) = P(C R) P(C R) = = P(R A)P(A) = P(R B)P(B) = P(R C)P(C) so Barb is most likely to have been responsible. =.6(.3).44 =.3(.7).44 =.1(.5).44 =.49 =.4772 = (a) If A and B are events, prove that P(A) = P(A B) + P(A B). Solution: Since S = B B, A = A S = A (B B) = (A B) (A B) where the last equality is one of DeMorgan s laws. Because B B = and A B B and A B B then A B and A B are disjoint. Thus,. P(A) = P ((A B) (A B)) = P(A B) + P(A B) (b) Assuming further that B A, use the above to prove that P(B) P(A). Solution: If B A then A B = B so P(A) = P(B) + P(A B) and since P(A B) because of the second axiom of probability then P(A) P(B).

4 7. The percentage of couples where both people work is 52.1%. In a randomly selected sample of 1 couples, what is the probability that at least half of those couples have both people working. Solution: For a binomial random variable X with n = 1 and p =.521, P(X 5) = 1 P(X 4) = Consider the following data from a classic work of statistics. Ladislaus von Bortkiewicz collected data on how many Prussian cavalry soldiers were fatally kicked by horses in 2 cavalry troops over 1 years. For the 2 observations, he recorded how many people had died in that troop in that year. His observations can be summarized in the following table Number of Deaths Occurrences (a) Find the average number of people who died from horse kicks in a year in one troop of the Prussian cavalry. Solution: X = (19) + 1(65) + 2(22) + 3(3) + 4(1) = 61 1 (b) Using this average as the parameter of a Poisson random variable, find the theoretical proportion of years that should have y deaths for y =, 1, 2, 3and4. Solution: p() = λ e λ! p(1) = (.61)1 e.61 1! p(2) = (.61)2 e.61 2! p(3) = (.61)3 e.61 3! p(4) = (.61)4 e.61 4! = e

5 (c) Does the Poisson distribution fit the observed data well? Solution: Yes. Consider the following table where the expected number of deaths is just 2p(j). The expected and observed numbers are close in every case. j Observed Expected The response times for an online mortgage quote have a gamma distribution with mean four seconds and variance eight seconds 2. Give the probability distribution for the response time and using Tchebysheff s theorem, find an interval that contains at least 9% of the response times. Solution: The gamma distribution has µ = αβ and σ = αβ 2 so setting αβ = 4 and αβ 2 = 8 and solving gives α = 2, β = 2 which gives a probability density function of f (y) = yα 1 e y/β β α Γ(α) = y2 1 e y/2 2 2 Γ(2) = ye y/2 4 for positive values of y and f (y) = otherwise. Note that Γ(2) = (2 1)! = 1. Using Tchebysheff s theorem, with 1 1/k 2 =.9 k = 1 gives P(4 1 8 Y ).9 but since we know that P(Y ) =, we can say that P( Y 12.94).9 1. The Kumaraswamy distribution, like the beta distribution, is supported on the interval [, 1]. The probability density function is abx f (x) = a 1 (1 x a ) b 1 x 1 otherw ise By integrating the PDF, find the cumulative distribution function. (Hint: try u = x a.)

6 Solution: F(x) = = = x x x a f (t) dt abt a 1 (1 t a ) b 1 dt b(1 u) b 1 du where u = t a = (1 u) b x a = (1 x a ) ( 1) = 1 (1 x a ) 11. The scores on each component of the SAT are scaled so that they have a normal distribution with µ = 5 and σ = 1. Find the probability of scoring over 75 on one component of the SAT. Solution: By standardizing, P(Y > 75) = P (Z > 75 5 ) = P(Z > 2.5) Suppose Alice flips three fair coins and let X be the number of heads showing. Suppose Barbara flips five fair coins and let Y be the number of heads showing. If Z = X Y, find P(Z = z) for all possible values of z. Solution: Both X and Y have binomial distributions with p =.5 and n = 3 and n = 5 respectively. Form a grid /256 3/256 3/256 1/ /256 15/256 15/256 5/ /128 15/128 15/128 5/ /128 15/128 15/128 5/ /256 15/256 15/256 5/ /256 3/256 3/256 1/256 where the (i, j) cell gives P(Y = i, X = j) = p Y (i)p X (j). Then, we can sum along the diagonals to get the probabilities for each value of Z. For example, P(Z = ) = P(X =, Y = ) + P(X = 1, Y = 1) + P(X = 2, Y = 2) + P(x = 3, Y = 3) = 1/ / / /128 = 56/256. Doing this for all of the diagonals gives

7 z p(z) 1/256 8/256 28/256 56/256 7/256 56/256 28/256 8/256 1/256 and just to check p(z) = = = 1

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