Question Points Score Total: 137

Transcription

1 Math 447 Test 1 SOLUTIONS Fall 2015 No books, no notes, only SOA-approved calculators. true/false or fill-in-the-blank question. You must show work, unless the question is a Name: Section: Question Points Score Total: 137

2 1. On today s episode of the Monty Hall game show, you are shown seven cards, all face-down, of which one is the Q and six are black spot cards. The objective is to select the Q, which wins $100. The host lets you select a face-down card, and then shows you, from the other six face-down cards, two black spot cards, which he can always do, because he knows where all the cards are. He offers then you the opportunity to switch your choice to one of the four remaining face-down cards. We will suppose in what follows that you decide to switch your choice. Let G denote the event that your initial selection was the Q. Let W denote the event that your new selection (after switching) is the Q. (a) (3 points) Find P (Ḡ), P (W Ḡ), and P (W G). P (Ḡ) = 6, P (W Ḡ) = 1, and P (W G) = (b) (4 points) Write an equation for P (W ) using the Law of Total Probability. P (W ) = P (W Ḡ)P (Ḡ) + P (W G)P (G) (c) (4 points) Find P (W ). P (W ) = = (d) (4 points) After you have switched your choice, the host offers you the opportunity to see another black spot card. There will be no opportunity to switch choices again. In these circumstances, how much are you willing to pay to see another black spot card? Why? Nothing, because seeing the card after making your choice does not change the probability that your selection was the right one. 2. We know the following about a colormetric method used to test lake water for nitrates. If water specimens contain nitrates, a solution dropped into the water will cause the specimen to turn red 95% of the time. When used on water specimens without nitrates, the solution causes the water to turn red 10% of the time (because chemicals other than nitrates are sometimes present and they also react to the agent). Past experience in a lab indicates that nitrates are contained in 30% of the water specimens that are sent to the lab for testing. If a water specimen is randomly selected and turns red when tested, what is the probability that it actually contains nitrates? Solve this problem according to the following scheme. Make sure to give an appropriate answer for each part. (a) (3 points) Establish notation: name the relevant events. Let R denote the event that the specimen turns red and N denote the event that the specimen contains nitrates. (b) (3 points) In the notation of part (a), what information does the problem give you? P (R N) = 0.95 P (N) = 0.3 P (R N) = 0.1 Page 2

3 (c) (2 points) In the notation of part (a), what probability does the problem ask you to compute? P (N R) (d) (3 points) Write down the equation that lets you compute this answer, again in the notation of part (a). Bayes Formula: P (N R) = P (R N)P (N) P (R N)P (N) + P (R N)P ( N) (e) (3 points) Solve the problem, giving your answer as a percentage..95(.3) %.95(.3) +.1(.7) 3. (6 points) Before going on vacation for a week, you ask your absent-minded friend to water your ailing plant. Without water, the plant has a 90 percent chance of dying. Even with proper watering, it has a 20 percent chance of dying. And the probability that your friend will forget to water it is 30 percent. If your friend forgets to water the plant, the probability it will be dead when you return is closest to: (a) 0.93 (b) 0.83 (c) 0.73 (d) 0.63 (a) The answer is 90%; this is given to you in the problem and does not require the use of any formula at all. 4. Define a random variable Y such that p( 1) = 1 50 p(0) = (a) (3 points) Find E[Y ] Zero. E[Y ] = = 0. p(1) = 1 50 (b) (3 points) Find V [Y ] V [Y ] = E[Y 2 ] E[Y ] 2 = = (c) (3 points) Fill in the blank in the following statement of Tchebysheff s inequality: 1 k 2 P ( Y µ kσ) Page 3

4 (d) (2 points) What are µ and σ for the Y of this problem? Give numerical answers. µ = 0, σ = 1 5 (e) (2 points) For the Y of this problem, what bound does Tchebysheff s inequality give you for the probability P ( Y µ 1)? Use k = 5, and get P ( Y µ 1) 1 25? (f) (2 points) Again for the Y of this problem, what is P ( Y µ 1)? P ( Y 1) = = 1. Note that this shows the inequality cannot be improved, because we got the same number in the previous part of the problem. 5. Let X be a geometric random variable with parameter p. Write q for 1 p. All your answers below should be given in terms of p and q. (a) (4 points) Give a formula for p(k), where k is a positive integer. p(k) = q k 1 p. (b) (4 points) Find P (X > 2). P (X > 2) = 1 p(0) p(1) = 1 p qp = q qp = q(1 p) = q 2. (c) (4 points) Find P (X > 5 X > 3). The answer again is q 2, because asking that X > 5 if we already have X > 3 is the same as asking for two more Bernoulli trials until a success. Since the trials are independent, the failures on the previous trials don t affect the probability. Of course the same result can be obtained by summing series and using the definition of conditional probability. (d) (4 points) The geometric distribution has a memoryless property, indicated by the computations of the previous parts. For positive integers a and b, find P (X > a + b X > b). You may either compute, or guess, so no work is required. The answer is q a, by the reasoning of the previous part. 6. A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has four identical components, each with a probability of 0.2 of failing in less than 1000 hours. The subsystem will operate if any two of the four components are operating. Assume that the components operate independently. (a) (4 points) Find the probability that exactly two of the four components last longer than 1000 hours. ( ) 4 (0.2) 2 (0.8) 4 2 = 15.36% 2 (b) (6 points) Find the probability that the subsystem operates longer than 1000 hours. Page 4

5 ( ) 4 (0.2) 2 (0.8) ( ) 4 (0.2) 1 (0.8) ( ) 4 (0.2) 0 (0.8) 4 0 = 97.28% 0 7. In this question you will find, with proof, the moment generating function for the Poisson random variable. (a) (3 points) Fill in the blank in the following definition of a Poisson random variable: A random variable Y has the Poisson distribution with parameter λ > 0 if p(y) = for y = 0, 1, 2,.... λ y e λ y! (b) (2 points) Fill in the blank in the following definition of moment generating function: the moment generating function of a random variable Y is e ty m Y (t) = E[ ] (c) (2 points) Express m Y (t) for the Poisson random variable as the sum of an infinite series. λ y e λ m Y (t) = e ty y! y=0 (d) (7 points) Sum the series to determine m Y (t). e λ(et 1), see Example 3.23 on p (e) (2 points) Find m Y (0). λ, because the mean of the variable is the first derivative of the moment generating function, evaluated at zero. (f) (2 points) Fill in the blank in the following statement: the parameter λ used to define the Poisson distribution happens also to be the of the random variable Y. mean ; see previous question. 8. Two dice are tossed. (You may assume these are standard 6-sided dice, independent of one another, and not loaded.) Page 5

6 (a) (2 points) What is the probability that the total shown is 3? There are 36 possibilities; (1, 2) and (2, 1) are the only ones with a total of 3, so the answer is 2/36 = 1/18. (b) (2 points) What is the probability that the total shown is 4? 3/36 = 1/12, similarly. (c) (7 points) If the dice are tossed repeatedly, what is the probability that we obtain a total of 3 before obtaining a total of 7? The probability of obtaining a 7 is 6/36 = 1/6. In any given roll, we reduce to 3 possibilities: a total of 3 (probability p = 1/18), a total of 7 (probability r = 1/6), and something else (probability s = 28/36). Obtaining a 3 before a 7 requires that we do so on roll n for some n. The probability of obtaining a 3 before a 7 on the first roll is p, and on the second roll sp, because to get to the second roll the first roll must be not a 3 and not a 7. Similarly, the probability of obtaining a 3 before a 7 on the n-th roll is s n 1 p. Summing, we get the probability s n 1 p = n=1 p 1 s = If two events, A and B, are such that P (A) =.5, P (B) =.3, and P (A B) =.1, find the following: (a) (2 points) P (A B) P (A B) = P (A B) P (B) = 1 3 (b) (2 points) P (B A) P (B A) = P (B A) P (A) = 1 5 (c) (3 points) P (A A B) Note that P (A B) = P (A) + P (B) P (A B) = 0.7. Also note that A (A B) = A. The proceed as before to get P (A A B) = P (A) P (A B) = In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five who had enrolled in paid internships and three who enrolled in traditional student teaching programs. All eight candidates appear to be equally qualified, so three are randomly selected to fill the open positions. Let Y be the number of internship-trained candidates who are hired. Page 6

7 (a) (2 points) Does Y have a binomial or hypergeometric distribution? Hypergeometric. (b) (2 points) Why? The selection is done without replacement, since a candidate cannot hold more than one job. Also, the selection pool is not so large relative to the sample that the binomial is a good approximation to the hypergeometric. (To get credit for this question, you need to give some reason that distinguishes the hypergeometric distribution from the binomial distribution. For example, the answer by definition of hypergeometric is not adequate.) (c) (4 points) Find the probability that two or more internship trained candidates are hired. Recall that ( r N r ) y)( p(y) = Here N = 8, r = 5, and n = 3. We are interested in p(2) + p(3), which is ( ) ( ) p(2) + p(3) = 2)( 3 2 ( 8 + 3)( 3 3 ) = 3) = 40/ % 56 ( 8 3 n y ( N n) (d) (2 points) What is the mean of Y? n r N = = 15 8 = (e) (2 points) What is the variance of Y? n r N r N n N N N 1 = A geological study indicates that an exploratory oil well should strike oil with probability 0.2. (a) (5 points) What is the probability that the third strike comes on the seventh well drilled? Using the probability function for the negative binomial distribution: ( ) 7 1 p 3 q 7 3 = 15(0.2) 3 (0.8) 4 = (b) (2 points) What assumption did you make to obtain the answer to part (a)? Independence (of the wells drilled striking oil). (c) (6 points) Continuing under this assumption, find the mean and variance of the number of wells that must be drilled if the company wants to set up three producing wells. µ = r/p, σ 2 = rq/p 2, so µ = 15 and σ 2 = 60. Page 7