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Chapter 4Introduction to ProbabilityLearning Objectives1.Obtain an appreciation of the role probability information plays in the decision making process.2.Understand probability as a numerical measure of the likelihood of occurrence.3.Know the three methods commonly used for assigning probabilities and understand when they should be used.4.Know how to use the laws that are available for computing the probabilities of events.5.Understand how new information can be used to revise initial (prior) probability estimates using Bayes theorem.4 - 1Introduction to ProbabilitySolutions:1.Number of experimental Outcomes = (3) (2) (4) = 242.ABCACEBCDBEFABDACFBCECDEABEADEBCFCDFABFADFBDECEFACDAEFBDFDEF3.BDF BFD DBF DFB FBD FDB4.a.b.Let: H be head and T be tail(H,H,H)(T,H,H)(H,H,T)(T,H,T)(H,T,H)(T,T,H)(H,T,T)(T,T,T)c.The outcomes are equally likely, so the probability of each outcomes is 1/8.5.P(Ei) = 1 / 5 for i = 1, 2, 3, 4, 5P(Ei) 0 for i = 1, 2, 3, 4, 5P(E1) + P(E2) + P(E3) + P(E4) + P(E5) = 1 / 5 + 1 / 5 + 1 / 5 + 1 / 5 + 1 / 5 = 1The classical method was used.6.P(E1) = .40, P(E2) = .26, P(E3) = .34The relative frequency method was used.7.No. Requirement (4.3) is not satisfied; the probabilities do not sum to 1. P(E1) + P(E2) + P(E3) + P(E4) = .10 + .15 + .40 + .20 = .858.a.There are four outcomes possible for this 2-step experiment; planning commission positive - council approves; planning commission positive - council disapproves; planning commission negative - council approves; planning commission negative - council disapproves.b.Letp = positive, n = negative, a = approves, and d = disapproves9.10.a.Use the relative frequency approach:P(California) = 1,434/2,374 = .60b.Number not from 4 states = 2,374 - 1,434 - 390 - 217 - 112 = 221P(Not from 4 States) = 221/2,374 = .09c.P(Not in Early Stages) = 1 - .22 = .78d.Estimate of number of Massachusetts companies in early stage of development - (.22)390 86e.If we assume the size of the awards did not differ by states, we can multiply the probability an award went to Colorado by the total venture funds disbursed to get an estimate.Estimate of Colorado funds = (112/2374)($32.4) = $1.53 billionAuthors Note: The actual amount going to Colorado was $1.74 billion.11.a.No, the probabilities do not sum to one. They sum to .85.b.Owner must revise the probabilities so they sum to 1.00.12.a.Use the counting rule for combinations:b.Very small: 1/1,906,884 = 0.0000005c.Multiply the answer to part (a) by 42 to get the number of choices for the six numbers.No. of Choices = (1,906,884)(42) = 80,089,128Probability of Winning = 1/80,089,128 = 0.000000012513.Initially a probability of .20 would be assigned if selection is equally likely. Data does not appear to confirm the belief of equal consumer preference. For example using the relative frequency method we would assign a probability of 5 / 100 = .05 to the design 1 outcome, .15 to design 2, .30 to design 3, .40 to design 4, and .10 to design 5.14.a.P (E2) = 1 / 4b.P(any 2 outcomes) = 1 / 4 + 1 / 4 = 1 / 2c.P(any 3 outcomes) = 1 / 4 + 1 / 4 + 1 / 4 = 3 / 415.a.S = ace of clubs, ace of diamonds, ace of hearts, ace of spadesb.S = 2 of clubs, 3 of clubs, . . . , 10 of clubs, J of clubs, Q of clubs, K of clubs, A of clubsc.There are 12; jack, queen, or king in each of the four suits.d.For a: 4 / 52 = 1 / 13 = .08For b: 13 / 52 = 1 / 4 = .25For c: 12 / 52 = .2316.a.(6) (6) = 36 sample pointsb.c.6 / 36 = 1 / 6d.10 / 36 = 5 / 18e.No. P(odd) = 18 / 36 = P(even) = 18 / 36 or 1 / 2 for both.f.Classical. A probability of 1 / 36 is assigned to each experimental outcome.17.a.(4, 6), (4, 7), (4 , 8)b.05 + .10 + .15 = .30c.(2, 8), (3, 8), (4, 8)d.05 + .05 + .15 = .25e.1518.a.0; probability is .05b.4, 5; probability is .10 + .10 = .20c.0, 1, 2; probability is .05 + .15 + .35 = .5519.a.Yes, the probabilities are all greater than or equal to zero and they sum to one.b.P(A) = P(0) + P(1) + P(2)= .08 + .18 + .32= .58c.P(B) = P(4) = .1220.a.P(N) =56/500 = .112b.P(T)=43/500 = .086c.Total in 6 states = 56 + 53 + 43 + 37 + 28 + 28 = 245P(B)=245/500 = .49Almost half the Fortune 500 companies are headquartered in these states.21.a.P(A) = P(1) + P(2) + P(3) + P(4) + P(5) = = .40 + .24 + .12 + .06 + .02= .84b.P(B) = P(3) + P(4) + P(5) = .12 + .06 + .02= .20c.P(2) = 12 / 50 = .2422.a.P(A) = .40, P(B) = .40, P(C) = .60b.P(A B) = P(E1, E2, E3, E4) = .80. Yes P(A B) = P(A) + P(B).c.Ac = E3, E4, E5 Cc = E1, E4 P(Ac) = .60 P(Cc) = .40d.A Bc = E1, E2, E5 P(A Bc) = .60e.P(B C) = P(E2, E3, E4, E5) = .8023.a.P(A)= P(E1) + P(E4) + P(E6) = .05 + .25 + .10 = .40P(B) = P(E2) + P(E4) + P(E7) = .20 + .25 + .05 = .50P(C) = P(E2) + P(E3) + P(E5) + P(E7) = .20 + .20 + .15 + .05 = .60b.A B = E1, E2, E4, E6, E7P(A B)= P(E1) + P(E2) + P(E4) + P(E6) + P(E7)= .05 + .20 + .25 + .10 + .05= .65c.A B = E4 P(A B) = P(E4) = .25d.Yes, they are mutually exclusive.e.Bc = E1, E3, E5, E6; P(Bc) = P(E1) + P(E3) + P(E5) + P(E6)= .05 + .20 + .15 + .10= .5024.P(Crash Not Likely) = 1 - .14 - .43 = .4325.LetY = high one-year returnM = high five-year returna.P(Y)=15/30 = .50P(M)=12/30 = .40P(Y M) = 6/30 = .20b.P(Y M)= P(Y) + P(M) - P(Y M)= .50 + .40 - .20 = .70c.1 - P(Y M) = 1 - .70 = .3026. LetY = high one-year returnM = high five-year returna.P(Y)=9/30 = .30P(M)=7/30 = .23b.P(Y M) = 5/30 = .17c.P(Y M)= .30 + .23 - .17 = .36P(Neither) = 1 - .36 = .6427.Let:D = consumes or serves domestic wineI= consumes or serves imported wineWe are given P(D) = 0.57, P(I) = 0.33, P(D I) = 0.63P(D I) = P(D) + P(I) - P(D I) = 0.57 + 0.33 - 0.63 = 0.2728.Let:B = rented a car for business reasonsP = rented a car for personal reasonsa.P(B P)= P(B) + P(P) - P(B P)= .54 + .458 - .30 = .698b.P(Neither) = 1 - .698 = .30229.a.b.A person can have only one primary cause of death listed on a death certificate. So, they are mutually exclusive.c.P(H C) = 0.299 + 0.222 = 0.521d.P(C S) = 0.222 + 0.066 = 0.288e.1 - 0.299 - 0.222 - 0.066 = 0.41330.a.b.c.No because P(A | B) P(A)31.a.P(A B) = 0b.c.No. P(A | B) P(A); the events, although mutually exclusive, are not independent.d.Mutually exclusive events are dependent.32.a.b.65% of the customers are under 30.c.The majority of customers are single: P(single) = .75.d.55e.Let:A = event under 30B = event singlef.P(A B) = .55P(A)P(B) = (.65)(.75) = .49Since P(A B) P(A)P(B), they cannot be independent events; or, since P(A | B) P(B), they cannot be independent.33.a.b.It is most likely a student will cite cost or convenience as the first reason - probability = .511. School quality is the first reason cited by the second largest number of students - probability = .426.c.P(Quality | full time) = .218 / .461 = .473d.P(Quality | part time) = .208 / .539 = .386e.For independence, we must have P(A)P(B) = P(A B).From the table, P(A B) = .218, P(A) = .461, P(B) = .426P(A)P(B) = (.461)(.426) = .196Since P(A)P(B) P(A B), the events are not independent.34.a.P(O) = 0.38 + 0.06 = 0.44b.P(Rh-) = 0.06 + 0.02 + 0.01 + 0.06 = 0.15c.P(both Rh-) = P(Rh-) P(Rh-) = (0.15)(0.15) = 0.0225d.P(both AB) = P(AB) P(AB) = (0.05)(0.05) = 0.0025e.f.P(Rh+) = 1 - P(Rh-) = 1 - 0.15 = 0.8535.a.P(Up for January) = 31 / 48 = 0.646b.P(Up for Year) = 36 / 48 = 0.75c.P(Up for Year Up for January) = 29 / 48 = 0.604P(Up for Year | Up for January) = 0.604 / 0.646 = 0.935d.They are not independent sinceP(Up for Year) P(Up for Year | Up for January)0.75 0.93536.a.Satisfaction ScoreOccupationUnder 5050-5960-6970-7980-89TotalCabinetmaker.000.050.100.075.025.250Lawyer.150.050.025.025.000.250Physical Therapist.000.125.050.025.050.250Systems Analyst.050.025.100.075.000.250Total.200.250.275.200.0751.000b.P(80s) = .075 (a marginal probability)c.P(80s | PT) = .050/.250 = .20 (a conditional probability)d.P(L) = .250 (a marginal probability)e.P(L Under 50) = .150 (a joint probability)f.P(Under 50 | L) = .150/.250 = .60 (a conditional probability)g.P(70 or higher) = .275 (Sum of marginal probabilities)37.a.P(A B)= P(A)P(B) = (.55)(.35) = .19b.P(A B) = P(A) + P(B) - P(A B) = .55 + .35 - .19 = .71c.P(shutdown) = 1 - P(A B) = 1 - .71 = .2938.a.b.This is an intersection of two events. It seems reasonable to assume the next two messages will be independent; we use the multiplication rule for independent events.P(E-mail Fax) = P(E-mail) P(Fax) = c.This is a union of two mutually exclusive events.P(Telephone Interoffice Mail) = P(Telephone) + P(Interoffice Mail) = 39.a.Yes, since P(A1 A2) = 0b.P(A1 B) = P(A1)P(B | A1) = .40(.20) = .08P(A2 B) = P(A2)P(B | A2) = .60(.05) = .03c.P(B) = P(A1 B) + P(A2 B) = .08 + .03 = .11d.40.a.P(B A1) = P(A1)P(B | A1) = (.20) (.50) = .10P(B A2) = P(A2)P(B | A2) = (.50) (.40) = .20P(B A3) = P(A3)P(B | A3) = (.30) (.30) = .09b.c.EventsP(Ai)P(B | Ai)P(Ai B)P(Ai | B)A1.20.50.10.26A2.50.40.20.51A3.30.30.09.231.00.391.0041.S1 = successful, S2 = not successful and B = request received for additional information.a.P(S1) = .50b.P(B | S1) = .75c.42.M = missed paymentD1 = customer defaultsD2 = customer does not defaultP(D1) = .05 P(D2) = .95 P(M | D2) = .2 P(M | D1) = 1a.b.Yes, the probability of default is greater than .20.43.Let:S = small carSc=other type of vehicleF=accident leads to fatality for vehicle occupantWe have P(S) = .18, so P(Sc) = .82. Also P(F | S) = .128 and P(F | Sc) = .05. Using the tabular form of Bayes Theorem provides:EventsPrior ProbabilitiesConditional ProbabilitiesJoint ProbabilitiesPosterior ProbabilitiesS .18.128.023 .36Sc .82.050.041 .64 1.00.0641.00From the posterior probability column, we have P(S | F) = .36. So, if an accident leads to a fatality, the probability a small car was involved is .36.44.LetA1 = Story about Basketball TeamA2 = Story about Hockey TeamW = We Win headlineP(A1) = .60P(W | A1) = .641P(A2) = .40P(W | A2) = .462AiP(Ai)P(W | A1)P(W Ai)P(Ai | M )A1.60.641.3846.3846/.5694= .6754A2.40.462.1848.1848/.5694= .3246.5694 1.0000The probability the story is about the basketball team is .6754.45.a. EventsP(Di)P(S1 | Di)P(Di S1)P(Di | S1)D1.60.15.090.2195D2.40.80.320.78051.00 P(S1) = .4101.000P(D1 | S1) = .2195P(D2 | S1) = .7805b.EventsP(Di)P(S2 | Di)P(Di S2)P(Di | S2)D1.60.10.060.500D2.40.15.060.5001.00 P(S2) = .1201.000P(D1 | S2) = .50P(D2 | S2) = .50c.EventsP(Di)P(S3 | Di)P(Di S3)P(Di | S3)D1.60.15.090.8824D2.40.03.012.11761.00 P(S3) = .1021.0000P(D1 | S3) = .8824P(D2 | S3) = .1176d.Use the posterior probabilities from part (a) as the prior probabilities here.EventsP(Di)P(S2 | Di)P(Di S2)P(Di | S2)D1.2195.10.0220.1582D2.7805.15.1171.84181.0000.13911.0000P(D1 | S1 and S2) = .1582P(D2 | S1 and S2) = .841846.a.P(Excellent) = .18P(Pretty Good) = .50P(Pretty Good Excellent) = .18 + .50 = .68Note: Events are mutually exclusive since a person may only choose one rating.b.1035 (.05) = 51.75We estimate 52 respondents rated US companies poor.c.1035 (.01) = 10.35We estimate 10 respondents did not know or did not answer.47.a.(2) (2) = 4b.Lets = successfulu = unsuccessfulc.O = E1, E2M = E1, E3d.O M = E1, E2, E3e.O M = E1f.No; since O M has a sample point.48.a.P(satisfied) = 0.61b.The 18 - 34 year old group (64% satisfied) and the 65 and over group (70% satisfied).c.P(not satisfied) = 0.26 + 0.04 = 0.3049.LetI = treatment-caused injuryD= death from injuryN=injury caused by negligenceM=malpractice claim filed$=payment made in claimWe are given P(I) = 0.04, P(N | I) = 0.25, P(D | I) = 1/7, P(M | N) = 1/7.5 = 0.1333, and P($ | M) = 0.50a.P(N)=P(N | I) P(I) + P(N | Ic) P(Ic) = (0.25)(0.04) + (0)(0.96)=0.01b. P(D)=P(D | I) P(I) + P(D | Ic) P(Ic) = (1/7)(0.04) + (0)(0.96)=0.006c.P(M)=P(M | N) P(N) + P(M | Nc) P(Nc) = (0.1333)(0.01) + (0)(0.99)=0.001333P($)=P($ | M) P(M) + P($ | Mc) P(Mc) = (0.5)(0.001333) + (0)(0.9987)=0.0006750.a.Probability of the event= P(average) + P(above average) + P(excellent)= = .22 + .28 + .26= .76b.Probability of the event= P(poor) + P(below average)= 51.a.P(leases 1)=168 / 932 = 0.18b.P(2 or fewer)=401 / 932 + 242 / 932 + 65 / 932 = 708 / 932 = 0.76c.P(3 or more)=186 / 932 + 112 / 932 = 298 / 932 = 0.32d.P(no cars)=19 / 932 = 0.02 52.a.b.2022c.2245 + .1283 + .1090 = .4618d.400553.a.P(24 to 26 | Yes) = .1482 / .4005 = .3700b.P(Yes | 36 and over) = .0253 / .1090 = .2321c.1026 + .1482 + .1878 + .0917 + .0327 + .0253 = .5883d.P(31 or more | No) = (.0956 + .0837) / .5995 = .2991e.No, because the conditional probabilities do not all equal the marginal probabilities. For instance, P(24 to 26 | Yes) = .3700 P(24 to 26) = .336054.LetI=important or very importantM=maleF=femalea.P(I) = .49 (a marginal probability)b.P(I | M) = .22/.50 = .44 (a conditional probability)c.P(I | F) = .27/.50 = .54 (a conditional probability)d.It is not independentP(I) = .49 P(I | M) = .44and P(I) = .49 P(I | F) = .54e.Since level of importance is dependent on gender, we conclude that male and female respondents have different attitudes toward risk.55.a.We have P(B | S) P(B).Yes, continue the ad since it increases the probability of a purchase.b.Estimate the companys market share at 20%. Continuing the advertisement should increase the market share since P(B | S) = .30.c.The second ad has a bigger effect.56.a.P(A) = 200/800 = .25b.P(B) = 100/800 = .125c.P(A B) = 10/800 = .0125d.P(A | B) = P(A B) / P(B) = .0125 / .125 = .10e.No, P(A | B) P(A) = .2557.LetA = lost time accident in current yearB = lost time accident previous yearGiven: P(B) = .06, P(A) = .05, P(A | B) = .15a.P(A B) = P(A | B)P(B) = .15(.06) = .009b.P(A B)= P(A) + P(B) - P(A B)= .06 + .05 - .009 = .101 or 10.1%58.Let:A = return is fraudulentB = exceeds IRS standard for deductionsGiven: P(A | B) = .20, P(A | Bc) = .02, P(B) = .08, find P(A) = .3. Note P(Bc) = 1 - P(B) = .92P(A)= P(A B) + P(A Bc)= P(B)P(A | B) + P(Bc)P(A | Bc)= (.08)(.20) + (.92)(.02) = .0344We estimate 3.44% will be fraudulent.59.a.P(Oil) = .50 + .20 = .70b.Let S = Soil test resultsEventsP(Ai)P(S | Ai)P(Ai S)P(Ai | S)High Quality (A1).50.20.10.31Medium Quality (A2).20.80.16.50No Oil (A3).30.20.06.191.00 P(S) = .321.00P(Oil) = .81 which is good; however, probabilities now favor medium quality rather than high quality oil.60.a.A1 = field will produce oilA2 = field will not produce oilW = well produces oilEventsP(Ai)P(Wc | Ai)P(Wc Ai)P(Ai | Wc)Oil in Field.25.20.05.0625No Oil in Field.751.00.75.93751.00.801.0000The probability the field will produce oil given a well comes up dry is .0625.b.EventsP(Ai)P(Wc | Ai)P(Wc Ai)P(Ai | Wc)Oil in Field.0625.20.0125.0132No Oil in Field.93751.00.9375.98681.0000.95001.0000The probability the well will produce oil drops further to .0132.c.Suppose a third well comes up dry. The probabilities are revised as follows:EventsP(Ai)P(Wc | Ai)P(Wc Ai)P(Ai | Wc)Oil in Field.0132.20.0026.0026Incorrect Adjustment.98681.00.9868.99741.0000.98941.0000Stop drilling and abandon field if three consecutive wells come up dry.4 - 17
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