Practice Problems #4 PRACTICE PROBLEMS FOR HOMEWORK 4 (1) Read section 2.5 of the text. (2) Solve the practice problems below. (3) Open Homework Assignment #4, solve the problems, and submit multiple-choice answers. In each exercise, use the appropriate distribution. Binomial and Poisson probabilities can be computed directly by their PMF formulas or by using distribution tables, such as Tables C.1 and C.2 on pp. 781-788 of the text. 1. (10 marks) Ten certain supplier supplier are defective. defective. What is marks) Ten percent of computer parts produced by a certain the probability probability that a sample sample of 10 parts contains more than 3 defective defective ones? 2. (10 marks) On marks) On the average, two tornadoes hit major U.S. metropolitan areas every year. What is the probability that more than five tornadoes occur in major U.S. metropolitan areas next year? (As you probably saw in the news, a real tornado occurred near downtown Dallas on Sep 8, 2010.)
3. (10 marks) A marks) A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. probability that the virus enters at least 10 computers. computers. a) Find the probability b) A computer manager checks the lab computers, one after another, to see if they were infected by the virus. What is the probability that she has to test at least 6 computers to find the first infected one? 4. (10 marks) On marks) On the average, 1 computer in 800 crashes during a severe thunderstorm. A certain company had 4,000 working computers when the areawas hit by a severe thunderstorm. a) Compute the expected value and variance of the number of crashed computers. b) Compute the probability that less than 10 computers crashed. c) Compute the probability that exactly 10 computers crashed. (You may use a suitable approximation.)
5. (10 marks) A marks) A baker put 500 raisins into dough, mixed well, and made 100 cookies. You take a random cookie. cookie. What is the probabilit probability y of finding at least 4 raisins in it? (Hint: (Hint: use Poisson Poisson approximati approximation on to the Binomial distribut distribution ion of the number of raisins raisins in one cookie.)
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6. (10 marks) (Sec 2.5, p. 91, #3) A mischievous student wants to break into a computer file, which is password-protected. Assume that there are n passwords only one of which is correct, and that the student tries possible passwords in a random order. Let N be the number of trials required to break into the file. Determine the pmf of a) if unsuccessful passwords are not eliminated from further selections, and b) if they are eliminated. c) If n = 10, what is the expected number of trials in each case (a) and (b)? 7. (10 marks) An internet search engine looks for a certain keywordin a sequence of independent web sites. It is believed that 20% of the sites contain this keyword. a) Let X be the number of websites visited until the first keywordis found. Find the distribution of X . b) Compute the expected value and the standard deviation of X . c) Out of the first 10 websites, let Y be the number of sites that contain the keyword. Find the distribution of Y . d) Compute the expected value and the standard deviation of Y . e) Compute the probability that at least 5 of the first 10 websites contain the keyword. f) Compute the probability that the search engine had to visit at least 5 sites inorder to find the first occurrence of a keyword. 8. (10 marks) Identical computer components are shipped in boxes of 5. About 15% of components have defects. Boxes are tested in a random order. a) What is the probability that a randomly selected box has only non-defective components? b) What is the probability that at least 8 of randomly selected 10 boxes have only non-defective components? c) What is the distribution of the number of boxes tested until a box without defective components is found? 9. (10 marks) A master file consists of 150,000 records. When a transaction file is run against the master file, approximately 12,000 records are updated.The records to be updated are assumed to be distributed uniformly over the master file. A program reads all the records one by one and updates those which are necessary to update. What is the probability that a) the first record is to be updated? b) at least 20 records are read before the first record to be updated is found? Generated February 9, 2011 using MuchLearning.org. Please note that updates to content and solutions on the live site are frequent. This material may be protected by copyright.
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c) exactly 20 records are read before the first record to be updated is found? 10. (10 marks) The Stanley Cup winner is determined in the final series between two teams. The first team to win 4 games wins the Cup. Suppose that Dallas Stars advance to the final series, and they have a probability of 0.55 to win each game, and the game results are independent of each other. Find the probability that a) Dallas Stars wins the Stanley Cup b) seven games are required to determine the Cup winner (Hint: Without loss of generality, you can assume that the series continues until 7 games are played, even if the Cup winner is determined earlier. This ”change of Stanley Cup rules” will not change the answer to the problem!)
11. (10 marks) (The previous problem continued...) Suppose that the series continues until Dallas Stars win 4 games, even if the other rival wins the Cup earlier. a) What is the expected number of games to be played? b) Find the probability that the series consists of less than 10 games. c) Find the probability that by the end of the series, the other rival wins at least 3 games. 12. (10 marks) Suppose that the number of inquiries arriving at a certain interactive system follows a Poisson distribution with arrival rate of 12 inquiries per minute. Find the probability of 10 inquiries arriving a) in a 1-minute interval; b) in a 3-minute interval. c. What is the expectation and the variance of the number of arrivals during each of these intervals?
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Solutions: 1. We need to find P (X > 3) , where X is the number of defective parts in a sample of 10 parts. This X is the number of ”successes” in 10 trials, therefore, it has Binomialdistribution with parameters n = 10 and p = 0.1. From the Table of Binomial distribution,
P (X > 3) = 1
− F (3) = 1 − 0.9872 = 0.0128
Or, by the formula of Binomial PMF,
=1
10
− (0.9) −
P (X > 3) = 1 P (0) P (1) P (2) P (3) 10 9 10 9 8 (10)(0.1)(0.9)9 ( )(0.1)2 (0.9)8 ( )(0.1)3 (0.9)7 = 0.0128 2 6
− − ·
−
−
− − · ·
2. We need to find P (X > 5), where X is the number of tornadoes in major U.S. metropolitan areas next year. Tornadoes are rare events, therefore, this X has Poisson distribution with parameter λ = 2 years 1 . −
From the Poisson Table, P (X > 5) = 1
− F (5) = 1 − 0.983 = 0.017
3. a. Let X be the number of computers entered by the virus. Each of the 20 computers is either entered or not, thus X is the number of ”successes” in n = 20 Bernoulli trials. Hence, X has Binomial distribution with n = 20 and p = 0.4. From the Table of Binomial distribution,
≥ 10) = 1 − P (X ≤ 9) = 1 − 0.7553 = 0.2447
P (X
b. We need to find P (Y 6), where Y is the number of computers tested until the first infected computer is found. This is the number of trials required to see the first success, therefore, Y has Geometric( p = 0.4) distribution. Using Geometric PMF (and geometric series),
≥
∞
(0.6) P (Y ≥ 6) = y =6
y −1
(0.4) =
(0.6)5 (0.4) = (0.6)5 = 0.0778 1 0.6
−
Another solution... The computer manager has to check at least 6 computers if the first five were not infected. The first five computers are not infected with probability (1 0.4)5 = 0.0778.
−
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4. Let X be the number of crashed computers. This is the number of ”successes” (crashed computers) out of 4,000 ”trials” (computers), with the probability of success 1/800. Thus, it has Binomial distribution with parameters n = 4, 000 (large) and p = 1/800 (small). a. Using Binomial distribution with n = 4, 000 and p = 1/800,
E (X ) = np = 5 and V ar(X ) = np(1
− p) = 5(1 − 1/800) = 4.994
This distribution is approximately Poisson with parameter λ = np = 5. Using the Table of Poisson distribution with parameter 5, b. P (X < 10) = F (9) = 0.968 c. P (X = 10) = F (10)
− F (9) = 0.986 − 0.968 = 0.018
5. The number of raisins X in one cookie is Binomial with n = 500 and p = 0.01. It is approximately Poisson with parameter np = 5. The probability of finding at least 4 raisins is
≥ 4) = 1 − F (3) = 1 − 0.265 = 0.735
P (X from the Poisson Table. 6.
a) N has Geometric distribution with p = 1/n because it is the number of independent Bernoulli trials until the first success, and each trial is a success (correct password) with probability 1/n. Therefore,
P (N = x) = (1/n)(1
− 1/n)
x−1
for x = 1, 2, 3,...
b) Here, the probability of success increases after each failure, therefore, the distribution of N is not Geometric. Since the order is random, the correct password has the same chance to appear first, second, etc., in a sequence of trials. Therefore, P (N = x) = 1/n for x = 1, 2,...,n (this is called Discrete Uniform distribution). c) In (a), the distribution is Geometric, so
E (X ) =
1 = n = 10 p
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In (b), the distribution is Discrete Uniform, so 10
1 10 · 11 = = 5.5 E (X ) = xP (x) = x x=1
10
2 10
·
Certainly, a smarter strategy (b) yields a smaller expected number of trials. 7. a) X is Geometric( p = 0.2). b) E (X ) = 1/p = 5 and Std(X ) = c) Y is Binomial(n = 10, p = 0.2). d) E (Y ) = np = 2 and Std(Y ) =
≥ 5) = (1 − p)
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np(1 − p) = √ 2 · 0.8 = 1.265
≥ 5) = 1 − F (4) = 1 − 0.9672 = 0.0328 .
e) From the Binomial Table, P (X f) P (Y
√ 1 − p/p = √ 0.8/0.2 = 4.47.
= 0.4096 .
8. a) In a given box, let X be the number of non-defective components. This X is Binomial, n = 5, p = 0.85. The probability of a box with five non-defective components is
P (X = 5) = (0.85)5 = 0.44 .
b) Now, let Y be the number of boxes with only non-defective components. This Y is also Binomial. Its parameters are n = 10 and p = 0.44, where p is calculated in (a).
P (Y
≥ 8) = P (8)+P (9)+P (10) = ( 102· 9 )(.44) (.56) +(10)(.44) (.56)+(.44) 8
2
9
10
= 0.0249
c) This is precisely the number of trials needed to get the firstsuccess. Therefore, the distribution is Geometric with p = 0.44. 9. a) 12, 000/150, 000 = 0.08
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b) Let X be the number of records read to find the first record tobe updated. Then X has the Geometric distribution with p = 0.08, and P (X 20) = (0.92)19 = 0.2051 (Use the Geometric PMF and compute the geometric series)
≥
c) P (X = 20) = (0.92)19(0.08) = 0.0164 10. a) Let the series continue till 7 games are played, and let X be the number of games won by Dallas Stars. X has Binomial distribution with parameters n = 7, p = 0.55.
P ( Dallas wins ) = P (X
≥ 4) = 1 − F (3) = 0.6083
b) LetY be the number of games won by Dallas among the first 6 games. Then Y has a Binomial distribution with parameters n = 6, p = 0.55.
}
P ( 7 games required ) = P (Y = 3) = F (3)
{
− F (2) = 0.3032
11. a) The number of games played until the first Dallas Stars victory is Geometric with parameter p = 0.55 and expected value E (X ) = 1/p = 20/11. Then, the expected number of games until the 4-th victory is
E (X 1 + X 2 + X 3 + X 4 ) = 4
20 11
=7
3 games 11
b. We need to find the probability that Dallas Stars wins at least 4 of the first 9 games, and it equals 0.8342 . Use Binomial(n = 9, p = 0.55) distribution. c. This is the probability that the series lasts at least 7 games,which means that 6 games weren’t enough, so Dallas had no more than 3 wins after 6 games. Using Binomial distribution with n = 6, p = 0.55, we find that
≤ 3) = 0.5585
P (X
12.
a) Let X be the number of inquiries in a 1-min interval. Then X is a Poisson random variable with the parameter 12.
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P (X = 10) =
e
−12
10
· 12
10!
= 0.1048
b. Let X be the number of inquiries in a 3-min interval. Then X is a Poisson random variable with the parameter 36.
P (X = 10) =
e
−36
10
· 36
= 0.0000002337 10! c. For Poisson distribution with parameter λ, we have E (X ) = V ar(X ) = λ . Therefore, for a 1-min interval, E (X ) = V ar(X ) = 12, and for a 3-min interval, E (X ) = V ar(X ) = 36.
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