Homework 4 - KEY. Jeff Brenion. June 16, Note: Many problems can be solved in more than one way; we present only a single solution here.

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1 Homework 4 - KEY Jeff Brenion June 16, 2004 Note: Many problems can be solved in more than one way; we present only a single solution here. 1 Problem 2-1 Since there can be anywhere from 0 to 4 aces, the range space of X, that is, the set of values that X can take, is{0, 1, 2, 3, 4}. Every poker hand is equally likely, so we can find our desired probabilities by simply counting the number of possible hands and the number of hands that have a given number of aces. To create a poker hand with k aces, we must choose k cards from the 4 aces, and 5 k cards from the 48 non-ace cards. Thus, ( 4 48 P (X = k = k( 5 k ( Problem 2-2 To find the mean and variance in this case, we simply need to apply the given formulas; no special tricks are necessary. E[X] = = 13 6 E[X 2 ] = = 7 V ar(x = E[X 2 ] E[X] 2 = =

2 3 Problem 2-15ab For part a, we know that the sum of all the probabilities must be 1. Thus, we have k k k 1 3 = k which implies k = 7. Now that we have k, we can find the mean and 8 variance in the usual manner. 4 Problem 3-4 E[X] = = 11 7 E[X 2 ] = = 3 V ar(x = E[X 2 ] E[X] 2 = = For part a, we can derive a value of Y for each possible value of X. Since the values of X have probabilities given, the probabilities for the corresponding values of Y will be identical; for instance, since X = 10 with probability 0.1, Y = 2000(12 10 = 4000 with probability 0.1. The full set of values is: (4000, 0.1(2000, 0.3(0, 0.4( 2000, 0.1( 4000, 0.1(else, 0 For part b, we can find the values in the standard manner. However, we can simplify the calculations for Y by taking advantage of the linear properties of mean and variance. E[X] = = 11.8 E[X 2 ] = = V ar(x = E[X 2 ] E[X] 2 = = 1.16 E[Y ] = 2000(12 E[X] = 2000( = 400 V ar(y = V ar(x = =

3 5 Problem 5-9 We can define the variable X to be equal to the number of the first test that fails; X is distributed geometrically with parameter p = Table 5-1 on page 124 gives probability functions for a number of distributions; we find that: P (X = 5 = = Problem 5-11 Since P (X = 5 = p 4 (1 p, the plot should be the graph of p 4 p 5 over the interval [0, 1]. To maximize P (X = 5, we look for critical points; thus, we take the derivative of p 4 p 5 and set it equal to zero. We find that 4p 3 5p 4 = 0, which has solutions p = 0 and p = 0.8. p = 0 is obviously not the solution, and analysis of the second derivative shows that p = 0.8 is indeed a maximum. This also makes some amount of intuitive sense; if 4 out of 5 trials succeed, it seems more likely that the 5th trial is the first to fail. 7 Problem 5-12 For part a, we can define the variable X to be the number of the trip that a sale occurs on; X is distributed geometrically with parameter p = 0.1. Table 5-1 on page 124 gives the mean for a geometric random variable as 1 p, from which it follows that we expect a sale after 10 attempts. Thus, it will cost = $46000 to make a sale. For part b, since we expect to spend $46,000 and will only profit $15,000, the trips should not be taken. For part c, we wish to know P (X > 20, since we can afford to make 20 trips. An easy way to do this is to find the probability of 20 failures, which is simply , or Problem 5-34 For this problem, we define the random variable X to be the number of orders. 3

4 For part a, we wish to know P (X > 3. The easiest way to do this is to look up our parameters in the Appendix, Table I, where we find that P (X 3 = Thus, the probability that the journey must be made is = For part b, the expected value of X is simply λ = 2, again by Table 5-1. For part c, we wish to find n such that P (X n > 0.9. Again, checking the table in the Appendix gives us n = 4. For part d, we cannot simply invoke Table 5-1 again. Any situation where there would have been a demand for 4 or more parts must be counted as 3, since only three crews can be serviced in that event. Thus, we have: E[services] = 0 P (X = P (X = P (X = P (X 3 = = For part e, since we expect to have 2 requests per day, and expect to service at the tool crib, it follow that the expected number of journeys per day is their difference, or Problem 4A We can consider X to be the sum of n Bernoulli random variables, each with parameter p, since we are observing each individual event and then adding 1 to our total for each success. Again by Table 5-1, the variance of each Bernoulli random variable is p(1 p, so their sum is np(1 p. 10 Problem 4B The easiest way to find the probability mass function for X is to treat this as a counting problem. There are 2 k 1 possible sequences of flips, since the last flip must be a head. Thus, we must assign the remaining n 1 heads to the remaining k 1 flips, which gives us: ( k 1 n 1 P (X = k = 2 k 1 To find E[X], we can simply find E[X i ] where X i is the expected number of flips to get the i-th head after the (i 1st head is observed, and then sum 4

5 the E[X i ]s. Each E[X i ] is a geometric random variable with p = 0.5, so we expect each one to take 2 flips. Since there are n such E[X i ]s, their sum will be 2n. 11 Problem 4C In both cases, we must sum P (X = 0, P (X = 1, P (X = 2, and P (X = 3. For the binomial random variable, we have ( 100 P (X = k = (.01 k ( k k The poisson random variable has probability P (X = k = 1 e k! Summing up the probabilities for each one gives us and respectively, which are indeed very close. 5

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