STAT 418HW7 SolutionsSections 4.7- 4.951, 53, 54, 59, 61, 77, 78, 79; TE 16, 1751. Let X = {the number of typographical errors on a page}. It is reasonable to assumethat the number of words on a page, n, is large and that the probability of an error, p, issmall. We are given the expected number of errors per page, that is λ = np = .2, so we mayuse the Poisson distribution to approximate the requested probabilities.P (X = 0) =e−.2(.2)00!= e−.2= .8187P (X ≥ 2) = 1 − P (X = 1) − P (X = 0) = 1 − (.2)e−.2− e−.2= 1 − 1.2e−.2= .0175.53.a. The probability that both partners of an arbitrary couple were both born on April 30is, assuming independence and an equal chance of having being born on any given date,p = 1/3652. Hence, the number of such couples is approximately Poisson with meannp = 80, 000/3652= .6. Therefore, the probability that at least one couple was bothborn on this date is approximately 1 − e−.6= .4512.b. The probability that both partners of an arbitrary couple were born on the same dayof the year is p = 1/365. Hence, the number of such couples is approximately Poissonwith mean np = 80, 000/365 = 219.18. Hence, the probability of at least one such pairis 1 − e−219.18≈ 1.54. The Poisson distribution may also be used to approximate the probability that anumber of events will occur in a certain period of time. Let X = {the number of carsabandoned on the highway in a week}. Given: λ = 2.2.a.P (X = 0) = e−2.2= .1108.b.P (X ≥ 2) = 1 − P (X = 0) − P (X = 1) = 1 − .1108 − 2.2e−2.2= .645459. Let X = {the number of prizes you win in the 50 lotteries}. Given: n = 50, p = 1/100,and therefore λ = np = 50/100 = .5. We may approximate the probabilities with the Poissondistribution:P (X ≥ 1) = 1 − P (X = 0) = 1 − e−.5= .3935P (X = 1) = e−.5(.5) = .3033P (X ≥ 2) = 1 − P (X = 0) − P (X = 1) = 1 − e−.5− e−.5(.5) = 1 − e−.5(1.5) = .0902.61. We may approximate the Binomial distribution with the Poisson distribution, whereλ = np = .0014 ∗ 1000 = 1.4.P (X ≥ 2) = 1 − P (X = 0) − P (X = 1) = 1 −e−1.41.400!−e−1.41.411!= 1 − 2.4e−1.4.177. Let E ={there are k matches in the left-hand box once the right-hand box is emptied}.E will occur if and only if the Nth choice of the right-hand box is made at the N + N -k trial.View the right-hand box as a success. By the negative binomial distribution, with p = 1/2,r = N, and n = 2N − k, we seeP (E) =2N − k − 1N − 1122N −k.As there is an equal probability that the left-hand box is emptied first and there are k matchesin the right-hand box, the probability that at the moment when the first box is emptied, theother box contains exactly k matches is2P (E) =2N − k − 1N − 1122N −k−1.78. A success is to choose 4 balls which exactly consist of 2 black balls and 2 white balls.The probability of success isp =424284=1835.Because the trials are independent, p is the same for each trial, and we stop with our firstsuccess, X = {the number of trials required until the first success} follows the Geometricdistribution:P (X = n) = (1 − p)n−1p =1 −1835n−11835=17n−135183579. X is a hypergeometric random variable with n = 10, N = 100, and m = 6.a.P (X = 0) =941010010= .5223b.P (X > 2) = 1 − P (X = 0) − P (X = 1) − P (X = 2)= 1 −9410+94961+9486210010= .012562TE 16.P (X = i)P (X = i − 1)=e−λλii!e−λλi−1(i−1)!=λii.ddiP (X=i)P (X=i−1)= −λi2< 0.ii. i ≤ λ :P (X=i)P (X=i−1)≥ 1 and strictly decreasing as i increases; therefore P (X = i) is strictlyincreasing as i increases. Because i only takes integer values, P (X = i) reaches its maxat the the largest integer not exceeding λ.iii. i > λ :P (X=i)P (X=i−1)< 1 and strictly decreasing as i increases; therefore P (X = i) is strictlydecreasing as i increases.TE 17.a. If X ∼ B(n, p) and λ = np, then P (X is even) = (1 + (1 − p − p)n)/2 by TE 15.ThereforeP (X is even) = (1 + (1 − 2p)n)/2 = (1 + (1 − 2λ/n)n)/2 −→n→∞(1 + e−2λ)/2b.e−λ+ eλ=∞Xn=0(−λ)nn!+∞Xn=0λnn!=∞Xn=0λn((−1)n+ 1)n!= 2∞Xn=0λ2n(2n)!, because (−1)n+ 1 = 0 when n is odd and = 2 when n is even.P (X is even) = P (X = 0) + P (X = 2) + P (X = 4) + . . .= e−λ∞Xn=0λ2n/(2n)!, because X ∼ P (λ)= e−λ(eλ+ e−λ)/2 = (1 +