Homework 6 (due November 4, 2009)


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1 Homework 6 (due November 4, 2009 Problem 1. On average, how many independent games of poker are required until a preassigned player is dealt a straight? Here we define a straight to be cards of consecutive values that are not all of the same suit: for example 3 of hearts, 4 of hearts, of spades, 6 of hearts, and 7 of hearts. Solution: We first define the random variable X to be equal to the number of games until Bob is dealt a straight. Since the probability that Bob is dealt a straight is equal to some number p (which we will calculate in a moment. Then X is a geometric random variable with parameter p. Thus on average we can expect that it would take E[X] = 1 games until Bob is dealt a straight. p ( 2 Now to compute p we first note that there are possible hands that Bob could be dealt, all occuring equally likely. To count the number of those hands that give a straight we first could have the that the lowest card in the hand is any number ace through 10 (we are allowing ace to be low or high. Thus 10 possible choices for the lowest number. Once the numbers in the hand are picked there are 4 ways of choosing the suits for those numbers. However, this includes the hands where all of the cards are the same suit, but there are only 4 of such hands for each fixed string of numbers thus we have that 4 4 is the number of different possible arrangements of suits once the numbers of the cards are picked. Thus we must have that: And the value of E[X] = p = 10(4 4 ( 2 = ( 2 = (4 4 Problem 2. Suppose that the average number of cars abandoned weekly on a certain highway is 2.2. Approximate the probability that there will be: (a no abbandoned cars in the next week; (b at least 2 abandoned cars in the next week. 1
2 Solution: We assume that since the number of cars on the highway in a certain week is very large and that the probablilty of one being abandoned is low, that we can approximate the number of abandoned cars in a week by a poisson random variable X. Since it is given that the average is 2.2 a week, then we assume that λ = E[X] = 2.2. Thus : (a P {X = 0} = e (b P {X 2} = 1 e e Problem 3. Suppose that the number of misprints occurring in on a page is a Poisson random variable with parameter λ = 3. (a Find the probability that 3 or more errors occur on page 12. Solution: Let X be the Poisson random variable mentioned in the problem. Then we seek: P {X 3} = 1 P {X = 0} P {X = 1} P {X = 2} = 1 e 3 3e e 3 = e (b Find the probability 3 or more errors occur on page 12 if it is known that at least one error occurred on page 12. Solution: Let Y be the event that at least one error occurs on page 12. Then we have want P {X 3 Y }. By the definition of conditional probability we have: P {X 3 Y } = P ({X 3} Y P (Y However. The event in the numerator of the right hand side of the above equation is actually just P {X 3} Since the event that there are at least 3 errors is contained in Y. Thus we have: P {X 3 Y } = P {X 3} P (Y = e 3 1 e Problem 4. If X is a Poisson random variable witht he parameter λ, show that: E[X n ] = λe[(x + 1 n 1 ]. 2
3 Then use this to compute E[X 3 ]. Solution: To show this we write the definition of the left hand side of the equation as: E[X n ] = i=1 i n λ λi e i!. Where I have eliminated the i = 0 term since 0 n = 0. Now I recognize that i i! = 1 (i 1!. Thus we have: E[X n ] = i n 1 e λ λ i (i 1! = (j+1 n λ λj+1 e j! i=1 j=0 = λ (j+1 n e j=0 λ λj j! = λe[(x+1n 1 ]. In this last string of equalities I have reindexed by letting j = i 1. Then I factored out a λ overall, and recognized what was left to be the expectation value for (X + 1 n 1. This establishes the equality that I set out to prove. Now to compute E[X 3 ] we have that: E[X 3 ] = λe[(x ] = λe[x 2 + 2X + 1] = λ(e[x 2 ] + 2E[X] + 1 = λ 2 (λ λ 2 + λ = λ 3 + 3λ 2 + λ. Here we used that E[X 2 ] = λ(λ + 1 for a Poisson r.v. with parameter λ. Problem. The suicide rate in a certain state is 1 suicide per 100, 000 inhabitants per month. (a Find the probability that in a city of 400, 000 inhabitants within this state, there will be 8 or more suicides in a given month. Solution: Let N(t is the random variable equal to the number of suicides in t months, then this is a this describes a Poisson process with λ = 4. This we have that the mass function is given by: 4t (4ti P (N(t = i = e i!. Thus the probability that we are interested in in this case is P (N(1 8. P (N(1 8 = 1 e 4 7 k=0 4 k k! = p (b Find the probability that there will be less then 8 suicides in this city in 2 months. 3
4 Similar to part one we are now looking for P (N(2 7 or: P (N(2 7 = e 8 7 k=0 8 k k! (c What is the probability that there will be at least 2 months during the year that will have 8 or more suicides? If we let X be the random variable that counts the months in which there are more then 8 suicides then by (a, X is a binomial random variable associated to the parameters (12, p where p is computed in (a. Then this question asks for P (X 2 = 1 P (X = 0 P (X = 1, and is thus given by: ( 12 P (X 2 = 1 p 0 (1 p 12 0 ( 12 1 p 1 (1 p 1 1 = 1 (1 p 12 12p(1 p (d Counting the present month as month number 1, what is the probability that the first month to have 8 or more suicides will be month number i, i 1? If Y is the number of months required until the first success occurring, where success is a month in which there are 8 or more suicides then Y is a geometric random variable associated to the parameter p where p is the same as that calculated in (a. Then we are interested in P (Y = i which is given by: P (Y = i = p(1 p i 1 =.011(1.011 i 1 Problem 6. Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger then the other and wins each game with probability.6, independent of the outcomes of the other games. Find the probability that the stronger team wins the series in exactly i games. Do it for i = 4,, 6, 7. Compare the probability that the stronger team wins with the probability that it would win a 2 out of 3 series. Also in the best of 7 series, compute the expected number of games player. 4
5 Solution: If we let Y be the random variable equal to the number of games required for stronger team to win overall then Y is a negative binomial random variable with ( i 1 the parameters (4,.6. Thus we want to computer P (Y = i = i 4 for i = 4,, 6, 7. P (Y = 4 = P (Y = = P (Y = 6 = P (Y = 7 = ( =.1296 ( = ( = ( = Thus the probability that they win overall is the sum of these 4 cases which is In the case where we are talking about a two out of three case we define Z as the random variable equal to the number of games it takes for the stronger team to get to 2 wins. This is again a negative binomial random variable associated to the parameters (2,.6 and thus we have that: P (Z = 2 = P (Z = 3 = ( =.36 ( =.288 This the overall probability that they win the entire 2 out of 3 match is the sum of these two cases which gives.648. Thus overall the stronger team has a better chance of winning if they play a first to 4 series. Problem 7. If X is a geometric random variable, show analytically that P {X = n + k X > n} = P {X = k}. Then, give a verbal argument IN COMPLETE SENTENCES, using the interpretation of a geometric random variable as to why the preceding equation is true.
6 Solution: To see this analytically we first use the definition of conditional probablity to write the left hand side as: P {X = n + k X > n} = P ({X = n + k}{x > n} P {X > n} In the case where k = 0 it is clear that both sides of the statement we want to prove are both 0 so assume that k is at least 1. In this case the event that X = n + k is contained in the event that X > n thus their intersection is just the smaller event and we have: P {X = n + k X > n} = P {X = n + k} P {X > n} = p(1 pn+k 1. i=n+1 p(1 pi 1 Here the last equality is obtained by means of the face that X is geometric. Now we can reindex and compute the sum in the denominator to get that the denominator is equal to (1 p n. (You should be able to show this by reindexing and computing the sum of a geometric series. This give: P {X = n + k X > n} = p(1 pn+k 1 (1 p n = p(1 p k 1 = P (X = k. The last equality again comes from the fact that X is a geometric random variable. The verbal argument goes as follows. If I think of X as representing the number of trials required to get one success where success occurs on each trial independently with probability p, then if I am conditioning on X > n then I want to think of starting at the point where I have performed n trials all which have been failures. Thus if I want to know the probability that X = n + k in this situation its as if I am starting over at 0 at the n nt trial, and thus the conditional probability should be the same as P {X = k} since the trials are independent. Problem 8. For a hypergeometric random variable, determine P {X = k + 1} P {X = k}. Solution: If we assume X is hyper geometric with parameters (n, N, m then we have: P {X = k + 1} P {X = k} = ( m ( N m k + 1 n (k + 1 ( N ( m n k ( N n ( N m = n k (m k(n k (k + 1(N + m n + k + 1 6
7 Problem 9. Suppose that a batch of 100 items contains 6 that are defective and 94 that are nondefective. If X is the number of defective items in a randomly drawn sample of 10 items from the batch, find: (a P {X = 0} (b P {X > 2}. Solution: X is a hypergeometric random variable associated to the parameters (N = 100, m = 6, n = 10. So: P {X = 0} = ( 6 ( ( P {X > 2} = 1 P {X = 0} P {X = 1} P {X = 2} ( 6 ( 94 ( 6 ( ( 100 ( =
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