STAT 418HW9 Solutions(pp. 195-196)63, 65, 71, 8063. Let N(t) = {the number of people who enter the casino in time t}. It is reasonable toassume that N (t) ∼ Poisson(λ). We are given that λ = 1/2 = .5, so we may use the Poissondistribution to approximate the requested probabilities.a.P (N (5) = 0) =e−.5∗5(.5 ∗ 5)00!= e−2.5= .0821b.P (N (5) ≥ 4) = 1 − P(N(5) = 0) − P (N(5) = 1) − P (N(5) = 2) − P (N (5) = 3)= 1 − e−2.5− (2.5)e−2.5−(2.5)2e−2.52−(2.5)3e−2.53!= .2424.65. This is a Binomial experiment. Let X = {the number of positive blood tests}. Wemay approximate the requested probabilities with the Poisson distribution, where λ = np =500/103= .5a.P (X > 0) = 1 − P (X = 0) = 1 − e−.5= .3935b.P (X > 1|X > 0) =1 − P (X = 0) − P (X = 1)P (X > 0)=1 − e−.5− .5 ∗ e−.51 − e−.5= .2293c. Given that Jones has the disease, more than one soldier has the disease if any of theremaining 499 have the disease. Let Y = {the number of positive tests for the remaining499 soldiers}.P (X > 1|Jones knowingly has the disease) = P (Y > 0) = 1−P (Y = 0) = 1−e−.499= .3927d. Following the same logic in part (c), there are 500 − i soldiers left who could have thedisease. Let Y = {the number of positive tests for the remaining 500 - i soldiers}.P (Y > 0) = 1 − P (Y = 0) = 1 − e(500−i)/100071. Because the games are independent and p = 12/38,a.26385= 0.15b.263831238= 0.1011180.a. Let X = {winnings}. The payoff is considered to be fair if EX = 0. In the examplegiven, EX = .25 ∗ 3 + .75 ∗ (−1) = 0. When the player selects 2 numbers, P2,2=20801979=.06, and F = {Fair Payoff} is such that,EX = 0 = .06 ∗ F + .94 ∗ (−1)F = .94/.06 = $15.67.b.Pn,k=nk80 − n20 − k8020=nk2080·1979· · · · ·20 − k + 180 − k + 1·6080 − k· · · · ·60 − n + k + 180 − n + 1.For n = 10:k P10,k060∗59∗58∗57∗56∗55∗54∗53∗52∗5180∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0458110120∗60∗59∗58∗57∗56∗55∗54∗53∗5280∗79∗78∗77∗76∗75∗74∗73∗72∗71= .1796210220∗19∗60∗59∗58∗57∗56∗55∗54∗5380∗79∗78∗77∗76∗75∗74∗73∗72∗71= .2953310320∗19∗18∗60∗59∗58∗57∗56∗55∗5480∗79∗78∗77∗76∗75∗74∗73∗72∗71= .2674410420∗19∗18∗17∗60∗59∗58∗57∗56∗5580∗79∗78∗77∗76∗75∗74∗73∗72∗71= .1473510520∗19∗18∗17∗16∗60∗59∗58∗57∗5680∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0514610620∗19∗18∗17∗16∗15∗60∗59∗58∗5780∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0115710720∗19∗18∗17∗16∗15∗14∗60∗59∗5880∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0016810820∗19∗18∗17∗16∗15∗14∗13∗60∗5980∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0001910920∗19∗18∗17∗16∗15∗14∗13∗12∗6080∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0000061020∗19∗18∗17∗16∗15∗14∗13∗12∗1180∗79∗78∗77∗76∗75∗74∗73∗72∗71= .0000001c. EX = (−1) ∗ (.0458 + .1796 + .2953 + .2674 + .1473) + .0514 + 17 ∗ .0115 + 179 ∗ .0016 +1299 ∗ .0001 + 2599 ∗ .000006 + 24999 ∗ .0000001 =