Notes for Math 324, Part 17

126 Notes for Math 324, Part 17

Chapter 17 Common discrete distributions 17.1 Binomial Consider an experiment consisting by a series of trials. The only possible outcomes of the trials are success and failure. The outcome of one trial is independent of the other trials. The probability of obtain a success is the same for all the trials. Let p be the probability of obtaining a success. A sequence of trials like this is called a Bernoulli sequence of trials. Let { 1 if the i th trial results in a success X i = 0 if the i th trial results in a failure The sequence of r.v. s X 1, X 2,... is called a Bernoulli sequence of r.v. s. Let S = n i=1 X i be the number of successes in the first n trials. Then, for each 0 k n, (17.1) P[S = k] = ( n k ) p k (1 p) n k. Here is the proof of (17.1). The probability of obtaining success, (k), success, failure, (n k), failure in precisely this order is p k (1 p) n k. However, the k successes and the n k failures can appear in any possible order. The number of ways to permutate k identical S s and n k identical F s is ( n k). Definition 17.1.1. A r.v. S has discrete binomial distribution with parameters n and p if ( ) n P[S = k] = p k (1 p) n k, for 0 k n. k Here n is a positive integer and 0 p 1. Theorem 17.1. Let S be a r.v. with a binomial distribution with parameters n and p. Then, E[S] = np and Var(S) = np(1 p). 127

128 CHAPTER 17. COMMON DISCRETE DISTRIBUTIONS Problem 17.1. (# 40, Sample Test) A small commuter plane has 30 seats. The probability that any particular passenger will not show up for a flight is 0.10, independent of other passengers. The airline sells 32 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available. Answer: 0.1564 Solution: Let S be the number of passengers who show up. X has a binomial distribution with n = 32 and p = 0.90. Then, P[more than 30 passengers show up] = P[S 31] = P[S = 31] + P[S = 32] = ( ) 32 31 (0.90) 31 (0.1) + ( 32) (0.90) 32 = 0.1564. Problem 17.2. (# 40, May 00) A company prices its hurricane insurance using the following assumptions: (i) In any calendar year, there can be at most one hurricane. (ii) In any calendar year, the probability of a hurricane is 0.05. (iii) The number of hurricanes in any calendar year is independent of the number of hurricanes in any other calendar year. Using the company s assumptions, calculate the probability that there are fewer than 3 hurricanes in a -year period. Answer: 0.92 Solution: Let S be the number of hurricanes in years. X has a binomial distribution with parameters n = and p = 0.05. We need to find P[S < 3] = P[S = 0] + P[S = 1] + P[S = 2] = ( ) 0 (0, 05) 0 (0.95) + ( ) 1 (0, 05) 1 (0.95) 19 + ( ) 2 (0, 05) 2 (0.95) 18 = 0.9243. Problem 17.3. (# 37, May 01) A tour operator has a bus that can accommodate tourists. The operator knows that tourists may not show up, so he sells 21 tickets. The probability that an individual tourist will not show up is 0.02, independent of all other tourists. Each ticket costs 50, and is non-refundable if a tourist fails to show up. If a tourist shows up and a seat is not available, the tour operator has to pay 100 (ticket cost + 50 penalty) to the tourist. What is the expected revenue of the tour operator? Answer: 985 Solution: Let S be the number of tourists which show up. X has a binomial distribution with parameters n = 21 and p = 0.98. The revenue of the tour operator is the r.v. Y = g(s), where { (21)(50) if S g(s) = (21)(50) 100 if S = 21 So, the expected revenue is E[Y ] = E[g(S)] = (21)(50)P[S ] + ((21)(50) 100)P [S = 21] = (21)(50)(1 P [S = 21]) + ((21)(50) 100)P [S = 21] = 1050 100P [S = 21] = 1050 100 ( 21) (0.98) 21 = 985. Problem 17.4. (# 27, November 01) A company establishes a fund of 1 from which it wants to pay an amount, C, to any of its employees who achieve a high performance level

17.1. BINOMIAL 129 during the coming year. Each employee has a 2% chance of achieving a high performance level during the coming year, independent of any other employee. Determine the maximum value of C for which the probability is less than 1% that the fund will be inadequate to cover all payments for high performance. Answer: 60 Solution: Let S be the number of employees who achieve a high performance level during the incoming year. S has a binomial distribution with parameters n = and p = 0.02. The company pays a total of CS. We want to find the maximum value of C such that P[CS > 1] < 0.01. This condition is equivalent to P[CS 1] 0.99. We have that So, C = 60. P [S = 0] = ( ) 0 (.98) = 0.6676 P [S 1] = ( ) 0 (.98) + ( ) 1 (0.02)(.98) 19 = 0.94901 P [S 2] = ( ) 0 (.98) + ( ) 1 (0.02)(.98) 19 + ( ) 2 (0.02) 2 (0.98) 18 = 0.9929 Example 17.1. Let X 1, X 2, X 3 be three independent continuous random variables each with density function { 2 x, if 0 < x < 2, f(x) = 0 else. What is the probability that exactly 2 of the 3 random variables exceeds 1? Solution: Let S be the number of r.v. s from X 1, X 2, X 3 which exceeds 1. S has a binomial distribution with n = 3 and p = P [X 1 1]. We have that p = P [X 1 1] = 2 1 Now, P[S = 2] = ( 3 2) (0.0858) 2 (1 0.0858) = 0.02. ( ) ( ) 2x x 2 2 2 x dx = = 2 1 2 + 1 2 2 = 0.0858. Problem 17.5. (# 23, May 01) A hospital receives 1/5 of its flu vaccine shipments from Company X and the remainder of its shipments from other companies. Each shipment contains a very large number of vaccine vials. For Company X s shipments, 10% of the vials are ineffective. For every other company, 2% of the vials are ineffective. The hospital tests 30 randomly selected vials from a shipment and finds that one vial is ineffective. What is the probability that this shipment came from Company X? Answer: 0.10 Solution: We apply Bayes theorem. Let B 1 be the event that the shipment came from Company X. Let B 2 be the event that the shipment did not come from Company X. Let A be the event that one vial in a sample of 30 is ineffective. By the Bayes theorem P[B 1 A] = P[B 1 ]P[A B 1 ] P[B 1 ]P[A B 1 ]+P[B 2 ]P[A B 2 ] We know that P[B 1 ] = 1 = 0.2 and P[B 5 2] = 4 = 0.8. We also have that 5 ( ) 30 P[A B 1 ] = P[Binom(n = 30, p = 0.1) = 1] = (0.1)(.9) 29 = 0.1413 1 1

130 CHAPTER 17. COMMON DISCRETE DISTRIBUTIONS and So, P[A B 2 ] = P[Binom(n = 30, p = 0.02) = 1] = P[B 1 A] = (0.2)(0.1413) (0.2)(0.1413)+(0.8)(0.334) = 0.0957. ( ) 30 (0.02)(0.98) 29 = 0.334. 1 17.2 Geometric Consider a sequence of Bernoulli trials. Let Y be the number of the trial at which the first success occurs. Then, P[Y = 1] = P [success] = p, P[Y = 2] = P [failure, success] = (1 p)p, P[Y = 3] = P [failure, failure, success] = (1 p) 2 p, P[Y = 4] = P [failure, failure, failure, success] = (1 p) 3 p In general, for each k = 1, 2,..., Pr[Y = k] = (1 p) k 1 p. Definition 17.2.1. A r.v. Y has a geometric distribution with parameter p, 0 p 1, if Pr[Y = k] = (1 p) k 1 p, k = 1, 2... Theorem 17.2. Let Y be a geometric distribution with parameter p. Then, E[Y ] = 1 p and Var(Y ) = 1 p p 2. Proof. Taking the derivative to 1 = 1 x k=0 xk 1, 1 < x < 1, we get (1 x) 2 1 < x < 1. So, E[Y ] = k(1 p) k 1 p p = k=1 = k=1 kxk 1, (1 (1 p)) 2 = 1 p. Q.E.D. Example 17.2. Let Y be the number of the throw of fair coin where the first head appears. Then, Y has a geometric distribution with p = 1. E[Y ] = 2 and Var(Y ) = 2. 2 Example 17.3. Suppose that an ordinary six sided die is rolled repeatedly, and the outcome is noted in each roll. Assume that all the faces are equally probable. Let Y be the number of throw at which the first six occurs. Then, Y has a geometric distribution with p = 1 6. E[Y ] = 6 and Var(Y ) = 30. Let S n be the number of successes in the first n Bernoulli trials, we have that P[Y k] = P [S k 1 = 0] = (1 p) k 1. This is so, because {Y k} is equivalent to need k trials or more to obtain a success. {S k 1 = 0} equivalent to not obtain any success at all in the first k 1 trials. Example 17.4. A certain basketball player hits a shot with probability 0.3. probability that he needs to throw 10 shots or more in order to get one shot? What is the

17.3. NEGATIVE BINOMIAL 131 Solution: We have to find P(Y 10) = P(S 9 = 0) = (0.7) 9. Let Y be a random variable Y with a geometric distribution, Then, for each k 1 For each k, n 1, So, P [Y k + n X k + 1] = P [Y k] = P [failure, (k 1), failure] = (1 p) k 1 P [Y k + n] P [Y k + 1] P [Y k + n 1 Y k] = P [Y n]. = (1 p)k+n 1 (1 p) k = (1 p) n 1 = P [Y n]. This formula says that if at the k 1 th trial we have not obtained a success, the probability that it will take an extra n or more trials to get a success is the same as the probability that it will take n or more trial to get a success for a Bernoulli sequences starting from the first trial. Having observed a roulette not giving red for a some trial does not change the probability of obtaining red in the next trials. Problem 17.6. (# 7, Sample Test) As part of the underwriting process for insurance, each prospective policyholder is tested for high blood pressure. Let Y represent the number of tests completed when the first person with high blood pressure is found. The expected value of Y is 12.5. Calculate the probability that the sixth person tested is the first one with high blood pressure. Answer: 0.053 Solution: Let Y be the number of test at which the first person with high blood pressure is found. Then, Y has a geometric distribution with parameter p = P [a person has blood pressure]. E[Y ] = 1 p = 12.5. So, p = 0.08. We need to find P [Y = 6] = (1 0.08) 5 (0.08) = 0.052. 17.3 Negative binomial Consider a sequence of Bernoulli trials. Let Y r the number of the trail at which the r th sucess occurs, then ( ) k 1 P[Y r = k] = p r (1 p) k r, k = r, r + 1,.... r 1 This is so, because when {Y r = k} in the first k 1 trials, we obtain r 1 sucesses and in the k th trial we obtain another sucess. So, ( ) ( ) k 1 k 1 P[Y r = k] = p r 1 (1 p) (k 1) (r 1) p = p r (1 p) k r. r 1 r 1 A r.v. Y r having such as probability mass function is called a negative binomial distribution. Theorem 17.3. E[Y r ] = r p and Var(Y r) = r(1 p) p 2

132 CHAPTER 17. COMMON DISCRETE DISTRIBUTIONS Let S n be the number of successes in the Bernoulli trials, we have that P[Y r k] = P (S k 1 r 1). This is so, because {Y r k} is equivalent to need k or more trials to obtain the r th success. This is the same as in the number of accumulated succeses until the k 1 th trial is less than r. This is the same as {S k 1 r 1}. Example 17.5. Suppose that an fair six face die is rolled repeatedly, and the outcome is noted in each roll. What is the probability that the third 6 occurs on the seventh roll? Assume that all the faces are equally probable. Solution: Let Y 3 the number of the trial at which the third six occurs. Y 3 has a negative binomial distribution. We have that ( ) ( ) 3 ( ) 4 6 1 5 P[Y 3 = 7] =. 2 6 6 Problem 17.7. (# 11, November 01) A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is 3. The number of accidents that occur in any given 5 month is independent of the number of accidents that occur in all other months. Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs. Answer: 0.29 Solution: Let S n be the number of months with accidents in the first n month. Let Y 4 be the number of the month when the fourth month with accidents occurs. We need to find P (Y 4 8). P [Y 4 8] = 1 P [Y 4 = 4] P [Y 4 = 5] P [Y 4 = 6] P [Y 4 = 7] 1 ( ) 3 3 (0.6) 4 (0.4) 0 ( ) 4 3 (0.6) 4 (0.4) 1 ( ) 5 3 (0.6) 4 (0.4) 2 ( 6 3) (0.6) 4 (0.4) 3 = 0.2897. Alternatively, using the relation between the negative binomial and the binomial, P[Y 4 8] = P [S 7 3] = P [S 7 = 0] + P [S 7 = 1] + P [S 7 = 2] + P [S 7 = 3] = ( ) 7 0 (0.6) 0 (.4) 7 + ( ) 7 1 (0.6) 1 (.4) 6 + ( ) 7 2 (0.6) 2 (.4) 5 + ( 7 3) (.06) 3 (.4) 4 = 0.2897. 17.4 Poisson Definition 17.4.1. A r.v. X has a Poisson distribution with parameter λ > 0, if for each integer k 0, In particular, λ λk P [X = k] = e k!. P [X = 0] = e λ, P[X = 1] = e λ λ, P [X = 2] = e λ λ2 2.

17.4. POISSON 133 Theorem 17.4. Let X be a r.v. with a Poisson distribution with parameter λ > 0, then E[X] = λ and Var(X) = λ. Problem 17.8. (# 24, May 00) An actuary has discovered that policyholders are three times as likely to file two claims as to file four claims. If the number of claims filed has a Poisson distribution, what is the variance of the number of claims filed? Answer: 2 Solution: Let X be the number of claims a poliholder files. X is a r.v. with a Poisson distribution with parameter λ > 0. We know that P [X = 2] = 3P [X = 4]. So, λ λ2 e 2 λ 2 = 4, and λ = 2. Hence, Var(X) = λ = 2. = 3e λ λ4 4! Problem 17.9. (# 23, November 00) A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and 10, 000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. What is the expected amount paid to the company under this policy during a one-year period? Answer: 7,231 Solution: Let X be the number of number of major snowstorms per year that shut down business. X is a r.v. with a Poisson distribution with parameter λ = 1.5. The payment of the insurance company is Y = g(x), where { 0 if X 1 g(x) = 1000(X 1) if X 2 So, E[Y ] = E[g(X)] = E[1000(X 1)I(X 2)] = E[1000(X 1)] E[1000(X 1)I(X 1)] = 1000(E[X] 1) + 1000P(X = 0) = 1000(1.5 1) + 1000 e 1.5 = 7231. Problem 17.10. (# 19, November 01) A baseball team has scheduled its opening game for April 1. If it rains on April 1, the game is postponed and will be played on the next day that it does not rain. The team purchases insurance against rain. The policy will pay 1000 for each day, up to 2 days, that the opening game is postponed. The insurance company determines that the number of consecutive days of rain beginning on April 1 is a Poisson random variable with mean 0.6. What is the standard deviation of the amount the insurance company will have to pay? Answer: 699 Solution: Let X be the number of consecutive days of rain beginning on April 1. X is a r.v. with a Poisson distribution with parameter λ = 0.6. The payment made by the insurance company is Y = g(x), where 0 if X = 0 g(x) = 1000 if X = 1 00 if X 2

134 CHAPTER 17. COMMON DISCRETE DISTRIBUTIONS So, E[Y ] = E[g(X)] = 1000P(X = 1) + 00P(X 2) = 1000P(X = 1) + 00(1 P(X = 0) P(X = 1)) = 1000 e 0.6 (0.6) + 00(1 e 0.6 e 0.6 (0.6)) = 573.0897 E[Y 2 ] = E[(g(X)) 2 ] = 1000 2 P(X = 1) + 00 2 P(X 2) = 1000 2 P(X = 1) + 00 2 (1 P(X = 0) P(X = 1)) = 1000 2 e 0.6 (0.6) + 00 2 (1 e 0.6 e 0.6 (0.6)) = 816892.51. So, the standard deviation of the amount the insurance company will have to pay is σ = E[Y 2 ] (E[Y ]) 2 = 816892.51 (573.0897) 2 = 689.9. 17.5 Hypergeometric Consider an urn with N white balls and M white balls, k are withdrawn from the urn without replacement. Let X be the number of white balls selected, then ( N )( M ) i k i P[X = i] = ) ( N+M k for 0 i N and 0 k i M. The distribution of this r.v. is called hypergeometric. Definition 17.5.1. A r.v. X has a hypergeometric distribution with parameters N, M and k if ( N )( M ) i k i P[X = i] = ) for max(k M, 0) i min(n, k) ( N+M k Theorem 17.5. Let X be a r.v. hypergeometric distribution with parameters N, M and k, then E[X] = kn. N+M 17.6 Problems 1. An analysis of auto accidents shows that one in four accidents results in an insurance claim. In a series of independent accidents, find the probability that the first accident resulting in an insurance claim is one of the first 3 accidents. 2. If X has a Poisson distribution of that 3P [X = 1] = P [X = 2]. Find P [X = 4]. 3. A manufacturer of soft drink bottles turns out defective with probability 0.10. Assume that every bottle turn out defective independently of the rest of the bottles produced. Find the probability that there are exactly 4 defective bottles among the next 10 bottles produced. 4. Thirty percent of the applicants for a position have advanced training in computer programming. Three jobs requiring advanced training are open. Find the probability that the third qualified applicant is found on the fifth interview, supposing the applicants are interviewed sequentially and at random.

17.6. PROBLEMS 135 5. The lifetime in hours of a certain kind of radio tube is a random variable having a probability density function given by { 100 if x > 100, x f(x) = 2 0 else, What is the probability that exactly 2 of 5 such tubes in radio set will have to be replaced within the first 150 hours of operation? 6. A commuter s drive to work includes 7 stoplights. Assume that probability that a light is red when the commuter reaches it is 0., and that the lights are far enough apart to operate independently. Let X be the number of red lights the commuter stops for. Find the probability that the commuter has to stop for 2 or more lights. Find the expected value of X. 7. A system of 50 components functions if 45 or more of their components functions properly. Find the probability that the system operates if the probability that a component operates properly is 0.85, and different components operate or not independently of the rest. 8. A box contains 35 gems, of which 10 are real diamonds and 25 are fake diamonds. Gems are randomly taking out of the box, one at a time without replacement. What is the probability that exactly 2 fakes are selected before the second real diamond is selected? 9. A ball is drawn from an urn containing 4 white and 2 black balls. After the ball is drawn, it is then replaced and another ball is drawn. This goes indefinitely. What is the probability that, of the first 8 balls drawn, exactly 3 are white? 10. A certain basketball player hits a shot with probability 0.3. What is the probability that he needs to throw 10 or more shots in order to get 4 hits? 11. In an initial screening of job applicants, a recruiter accepts, on the average, one-third of all applicants for further consideration. In reviewing a collection of job applications (independent of one another), find the probability that the first application acceptable for further consideration is one of the first three applications reviewed. 12. On a six question multiple choice test there are five possible answers of which one is correct and four incorrect. If a student guesses randomly and independently find the probability that he gets at least 2 questions right? 13. Suppose that an ordinary six sided fair die is rolled repeatedly, and the outcome is noted in each roll. What is the probability that the third 6 occurs on the seventh roll? 14. What is more probable to get 2 or more sixes in 6 throws of a fair die or to get or more sixes in 60 throws of a fair die? Find each of these probabilities. 15. If the probability of hitting a target is 1 5, and then ten shoots are fired independently, what is the probability of the target being hit at least twice?

136 CHAPTER 17. COMMON DISCRETE DISTRIBUTIONS 16. A basketball player shoots 10 shots and the probability of hitting is 0.4 in on each shot. What is the probability of hitting exactly eight shots? 17. If X is a binomial random variable with expected value 6 and variance 2.4, find P (X = 5). 18. A fair die is tossed until a 2 is obtained. If X is the number of trials required to obtain the first 2, what is the smallest value of x for which P (X x) 1/2. 19. A certain hunter hits a duck with probability 0.3. What is the probability that he needs to throw 10 or more shots in order to get 2 or more ducks?. A ball is drawn from an urn containing 4 white and 2 black balls. After the ball is drawn, it is then replaced and another ball is drawn. This goes indefinitely. What is the probability that, of the first 8 balls drawn, exactly 3 are white? 21. The lifetime in hours of a certain kind of radio tube is a random variable having a probability density function given by { 100 for x > 100 x f(x) = 2 0 else A radio set has 50 of such tubes. What is the expected number of tubes in the radio set which will be still working after 150 hours of operation? 22. A man pays $1 a throw to try to win a $3 doll. His probability of winning on each throw is 0.1. What is the probability that more than three throws will be required to win the doll? 23. If a student selects true or false at random on an examination, assuming independence among the answers, determine the probability that at most three questions must be answered at random to obtain the first correct answer. 24. A deck of 52 cards is dealt out. What is the probability that the second ace occurs on the fourteenth card? 25. An experiment consists of tossing a fair die until a 6 occurs four times. What is the probability that the process ends after exactly ten tosses with a 6 occurring on the ninth and tenth tosses? 26. Find the probability that the fifth head is observed on the tenth independent flip of an unbiased coin. 27. A fair coin is flipped 4 times. What is the probability that the fourth flip is a head, given that each of the first flips resulted in heads? 28. Suppose that in 4 child families, each child is equally likely to be a boy or a girl, independently of the others. Which would then be more common, 4 child families with 2 boys and 2 girls, or 4 child families with different numbers of boys and girls?

17.6. PROBLEMS 137 29. Suppose that an ordinary six sided die is rolled repeatedly, and the outcome is noted in each roll. What is the probability that the third 6 occurs on the seventh roll? Assume that all the faces are equally probable.