Homework 11 & 12 Solutions (Section 12.4)

Transcription

1 Homework & 2 Solutions Section 24 December 9, 20 Homework Formula for the mean: n n = k n k EX = x xp X = x Formula for the variance: varx = E[X µ 2 ] = x x µ 2 P X = x Another way to compute the variance: varx = EX 2 [EX] 2 Using the above formula for the mean, we'd have: EX 2 = x x 2 P X = x Formula for the standard deviation: σ = varx Exercise 4 Roll a fair die twice X is the random variable that gives the maximum of the two numbers Find the probability mass function of X Let's rst make a chart of the outcomes We have

2 max and so we have a table for the probability mass function as follows x P X = x /6 2 /6 5/6 4 7/6 5 9/6 6 /6 2 Exercise 6 An urn contains ve green, two blue, and three red Remove three without replacement Let X denote the number of red balls We need to nd the probability that there are 0,, 2, and red balls 2 Method The probability that there are no red balls pulled is The probability that exactly one red ball is pulled is = where the rst term gives the probability of pulling out the red as the rst ball, the second term as the second ball, and third term as the third ball Similarly, the probability that exactly two red balls are pulled is

3 And the probability that exactly three red balls are pulled is Method 2 Alternatively, the total combinations of three balls removed without replacement is 0 To get 0 red we have 7 combinations For exactly red we have 7 2 For exactly two red we have 7 2 For exactly three red we have Taking the quotients, we get the same values as above 2 Solution x P X = x 0 20/720 78/ /720 6/ Check Solution If we counted correctly, these numbers should add up From the nal solution or Method, we have = 720 Good From Method 2 we have = 20 Exercise 4a Let Y be a random variable that counts the number of trials until the rst heads shows up, where the probability of heads is p P Y = is the probability that it takes one coin ip to obtain a heads on the rst trial This is exactly p P Y = 2 is the probability that it takes two coin ips until a heads appears Thus the rst trial must come up a tails and the second comes up heads The two coin ips are independent events, so we multiply their probability We have p for the rst trial and p for the second We conclude P Y = 2 = pp Similarily, P Y = requires two ips of tails and then a ip of heads and so we have corresponding probabilities of p, p, and p The events are independent so we multiple to obtain P Y = = p 2 p Remark : One notices a pattern and part b of exercise 4 says that P Y = j = p j p Remark 2: Part c of exercise 4 claims the sum of these innite events is by dening an innite sum as a limit of partial sums

4 4 Exercise 22 Given the following probability mass function x P x Compute the mean: EX = = = 0285 Compute the variance We already computed EX, so let's compute EX 2 and use formula on the rst page We have EX 2 = = = 0565 Thus, varx = EX 2 [EX] 2 = = And so the standard deviation is σ = varx Looking back at our probability distribution function, we could say this answer makes sense and is at least on the right order of magnitude 5 Exercise 24 Let X be uniformly distributed on the set S = {, 2,,, n} where n is a positive integer In other words, X has the probability mass function P X = k = n for any k S For a concrete example of this problem, do Exercise 2 4

5 We have EX = x S xp X = x = = = n n kp X = k k= n k= k n n k k= = nn + n 2 = n + 2 and we also have EX 2 = x S x 2 P X = x = = n n k 2 P X = k k= n k= k 2 = nn + 2n + n 6 n + 2n + = 6 Then varx = EX 2 [EX] 2 = n + 2n + 6 = 2 2n2 + n + 2 = n2 2 6 Exercise 0 n n2 + 2n + 2 The exercise provides us with the probability mass function for independent random variables X and Y 5

6 k P X = k P Y = k Then we can compute EX and varx using the following table: P X = k k kp X = k k 2 k 2 P X = k EX = 04 EX 2 = 26 It follows from the table that varx = Similarly we can compute EY and vary using the following table: P Y = k k kp Y = k k 2 k 2 P Y = k EX = 05 EX 2 = 2825 It follows from the table that vary = Then we have EX + Y = EX + EY = = 075 and varx + Y = varx + vary = 5 6

7 2 Homework 2 Binomial distribution Number of independent trials is n, number of successes is k, where the probability of a success is p n P X = k = p k p n k k Also important is knowing the steps to computing lim + a n n n or similar See Exercise 70 for an example and page 7 for another example We have [ lim exp n ln + a ] [ = exp lim n n n ln + a ] n n And we have lim n ln + a = lim n n n But since ln + a y lim y = lim y y = lim y ln + a n n + ay a y 2 a + a y y 2 = a by an application of l'hospital's rule, we determine that lim + n a n = e a A useful fact to know is that x + y n = n k=0 2 Exercise 4 n x k y n k 2 k Toss a coin with probability of heads equal to 0 ve times Let X be the number of tails What is P X = 2 and P X? Coin tosses are independent, so we have P X = 2 = To compute P X it's easier to just compute the probability of the complement P X = 0 and remember to subtract that answer from We have P X = 0 = = 0 5 and so P X = 0 5 7

8 22 Exercise 8 Four green and six blue balls Draw a ball at random Note its color Replace it Repeat these steps four times Let X denote the total number of green balls you obtain Find the probability mass function of X Replacement means the trials are independent Thus we have the following table: k P X = k Exercise 40 Four red, seven green, two white Draw, note, replace Repeat four times Let X denote the number of red balls and Y the number of green balls Find P X + Y = 2* *The book erroneously writes P X + Y = 2 2 Method X + Y = 2 occurs when the number of red and green balls is exactly two The probability of being either a red or green ball is out of Replacement means the trials are independent We have 4 P X + Y = 2 = p k p 4 k = k = = Method 2 X + Y = 2 means the number of white balls is always two Let Z denote the number of white balls Then P X + Y = 2 = P Z = = =

9 2 Method 2 Uses The Multinomial Distribution Longer, Still Good P X + Y = 2 = P X = 0, Y = 2 + P X =, Y = + P X = 2, Y = 0 = P X = 0, Y = 2, Z = 2 + P X =, Y =, Z = 2 + P X = 2, Y = 0, Z = 2 = 4! ! ! 4 0!2!2!!!2! 2!0!2! = 4! !2! = = Exercise 42 Prevalence of in 00 0 individuals pooled and tested What is the probability that the no individual has the disease? The probability is Exercise 46 A true-false exam has 20 questions What is the expected number of correct answers if a student guesses the answers at random This is a binomial distribution with 20 trials The probability of a student getting a particular question right at random is 05 The expected value is ES n = np = = 0 26 Exercise 54 Urn contains six green, eight blue, and ten red balls Take, note, replace Repeat six times What is the probability that you sampled two of each color? This is a multinomial distribution Let N be the number of greens, N 2 be the number of blues, and N be the number of reds Then P N = a, N 2 = b, N = c = a + b + c! a!b!c! a b c Since we repeat six times, the sum must be 6 In this particular question a = b = c = 2 P N = 2, N 2 = 2, N = 2 = =

10 27 Exercise 58 Attached earlobe is caused by a pair of recessive genes aa Given a couple Aa and aa, what is the probability the child has an unattached earlobe? Writing out the possibilities, we have A a a As aa a Aa aa and conclude the probability is a half 28 Exercise 60 Hemophilia is sex-linked recessive We mix X hemo X and XY Among the two sons and daughters, what is the probability that one daughter is not a carrier, one daughter is a carrier, one son is hemophilic, and one son is not hemophilic First we write out the possibilities X hemo X X X hemo X XX Y X hemo Y XY Thus there is probability p = 05 that a daughter is a carrier and p = 05 that a son is hemophilic Because there are two daughters and consider the situation for exactly one is a carrier, then the binomial distribution tells us that 2 pexactly one daughter is a carrier = = 05 and similarly 2 pexactly one son is hemophilic = = 05 To have the two situations occuring together, we multiple and get pone daughter is carrier, one is not, one son is hemophilic, one is not = Exercise 62 Coin p = of heads Probability rst time appears is fth trial Answer: Exercise 68 Rolling a fair die until the rst time a or 2 appears Find the probability that the rst or 2 appears within the rst ve trials 0

11 The probability that a or 2 appears is Note that the probability that it appears within the rst ve trials is the complement to it not showing within the rst ve trials Thus 5 2 p or 2 doesn't appear within the rst ve trials = For the unbelievers, we write: p = [ = 2 ] = = 2 Exercise 70 Urn contains one black and n white balls Balls drawn one at a time randomly until the black ball is selected Each ball is replaced before the next ball is drawn Find the probability that at least n draws are needed What happens as n? The probability of drawing a black ball is n The event of needing at least n draws happens when white is drawn for the rst n draws Thus the probability is n n Taking the limit as n goes to innity, we get the constant e As a review, we recall that the limiting case is special and can be dealt with by rst rewriting lim n [ exp n n n = lim [ = exp n ln ] n ] n lim n ln n The limit on the inside of the exponentiation is of type 0 and can be dealt with by yet again rewriting lim n ln ln n = lim n n n n

12 Finally we get a limit of the time 0 0 We apply l'hospital's rule to the corresponding functions lim x ln x x L H = lim x x 2 x = x 2 and conclude lim n = exp [ ] = n n e See the beginning of Section 2 for the general case 22 Exercise 74 Urn contains one black and n white balls Balls drawn one at a time randomly until the black ball is selected Each ball is replaced before the next ball is drawn Find the probability that exactly k white balls will be drawn before the black one is if a each ball is replaced before the next ball is drawn and b balls are not replaced If the balls are replaced, we have n n n k n If the balls are not replaced, we have n 2 n n n 2 n k n k + 2 Exercise 80 n k = n Suppose X is a Poisson distributed with parameter λ = 02 Find P X < and nd P 2 X Using the formula, we have λ λ P X = = e!, λ λ2 P X = 2 = e 2!, and λ λ P X = = e! P X < = P X = + P X = 2 and P 2 X = P X = 2 + P X = 2

13 24 Exercise 86 Suppose the number of phone calls arriving at a switchboard per hour is Poisson distributed with mean calls per hour Find the probabilty that at least one phone call arrives between noon and PM Assuming that phone calls in dierent hours are independent of each other, nd the probability that no phon calls arrive between noon and 2 PM Note that the mean tells us the value of λ, so λ = Then the probability of at least one phone call in an hour time slot is P X = P X < = P X = 0 λ λ0 = e 0! = e The probability that no phone calls arrive in two hours, due to the independence is the product of probabilities We have P X = 0 = e and so the probability is e 2 = e 6 25 Exercise 94 Suppose X and Y are indepedent and Poisson with mean λ Given that X + Y = n, nd the probability that X = k for k = 0,,, n The sum of Poisson random variables with means λ and λ 2 is another Poisson distributed with parameter λ + λ 2 The book derives this on page 75 It makes use of 2 on page 7 In any case, we have P X = k X + Y = n = P X = k and X + Y = n P X + Y = n The numerator is the same as P X = k and Y = n k Since X and Y are independent, we have P X = k and Y = n k = P X = kp Y = n k λ λk λ λn k = e e k! n k! = e 2λ λ n k!n k! Because X + Y is Poisson with mean 2λ, we have 2λ 2λn P X + Y = n = e n!

14 and so P X = k X + Y = n = 2 n n! k!n k! = 2 n n k 26 Exercise 96 in 500 experience side eects Using the Poisson approximation, nd the probability that in a group of 200 people, at least person experiences side eects At rst, we have P S 200 = P S 200 = 0 To approximate this with a Poisson distribution, we have the parameter λ which is the average, to be the np = = 2 5 Note that the probability of experiencing a side eect is small, part of what allows us to approximate using a Poisson distribution P X = 0 = e λ In conclusion, we have P S 200 e 2/5 27 Exercise 98 About in 000 boys is aected by fragile X syndrome, a genetic disorder that causes learning diculties Find the probability that, in a group of 500 boys, nobody is aected by this disorder by a computing the exact probability and b using a Poisson approximation The exact probability is and the Poisson approximation is λ = = 5 e λ I just used a calculator for both 4