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AN

ASSIGNMENT

THE CONCEPT OF PROBABILITY

(EXAMPLE NO. 4.34 TO 4.50)

TITLE OF BOOK: BUSINESS STATISTICS

AUTHOR: KEN BLACK

EDITION: 5th

PUBLISHER: JOHN WILEY SONS INC. U.S.A.

FOR THE SUBJECT OF

QUANTITATIVE TECHNIQUES

IN MANAGEMENT

IS

SUBMITED TO: Dr HIREN J. PATEL

ASSISTANT AND PROFESSIOR

OF VMPIM

SUBMITED BY:

Sr. NO. NAME ROLL NO.1) PRAJAPATI ALAY V. 117

2) PATEL NEHA 88

3) PATEL KIRAN 824) PATEL RAJ 100

5) PATEL DHAVAL 67

MBA-1 DIVISION: B

DATE OF SUBMISSION: 28/11/2010

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4.34

Use the values in the contingency table to solve the equation given.

Variable2 D E

A 10 20B 15 5C 30 15

55 40 95

a) P(E) = 40/95 = .42105

b) P(B B D) = P(B) + P(D) - P(B C D)

= 20/95 + 55/95 - 15/95 = 60/95 = .63158

c) P(A C E) = 20/95 = .21053

d) P(BE) = 5/40 = .1250e) P(A B B) = P(A) + P(B) = 30/95 + 20/95 =

50/95 = .52632

f) P(B C C) = .0000 (mutually exclusive)

g) P(DC) = 30/45 = .66667

h) P(AB)= P(A C B) = .0000 = .0000 mutually exclusiveP(B) 20/95

i) P(A) = P(AD)??

30/95 = 10/95 ??

.31579 { .18182No, Variables 1 and 2 are not independent.

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4.35

Use the value in contingency table to solve the equations given.

D E F G

A 3 9 7 12 31

B 8 4 6 4 22

C 10 5 3 7 25

21 18 16 23 78

a) P(F C A) = 7/78 = .08974

b) P(AB) = P(A C B) = .0000 = .0000P(B) 22/78

c) P(B) = 22/78 = .28205

d) P(E C F) = .0000 Mutually Exclusive

e) P(DB) = 8/22 = .36364

f) P(BD) = 8/21 = .38095g) P(D B C) = 21/78 + 25/78 10/78 = 36/78 = .4615

h) P(F) = 16/78 = .20513

4.36

The following probability matrix contains a breakdown on the age and gender of U.S. physicians

in a recent year as reported by the American medical association.

Age(years)

65

Gender Male .11 .20 .19 .12 .16 .78

Female .07 .08 .04 .02 .01 .22

.18 .28 .23 .14 .17 1.00

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a) what is the probability that one randomly selected physician is 35-44 yearsold?

P(35-44) = .28

b)

what is the probability that one randomly selected physician is both a womanand 45-54 years old?

P(Woman C 45-54) = .04

c) what is the probability that one randomly selected physician is a man or is 35-44 years old?

P(Man B 35-44) = P(Man) + P(35-44) - P(Man C 35-44) = .78 + .28 - .20 =

.86d) what is the probability that one randomly selected physician is less than 35

years old or 55-64 years old?

P(

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.78 + .40 - .27 = .91

b) P(NT C NK) = 1 - P(T B K) = 1 - .91 = .09

c) P(KT) = P(KC T)/P(T) = .27/.78 = .3462d) P(NT C K) = P(NT) - P(NT C NK)

but P(NT) = 1 - P(T) = .22

P(NT C K) = .22 - .09 = .13

4.38

The U.S. bureau of labor statistics publishes data on the benefits offered by small companies totheir employees. Only 42% offer retirement plans while 61% offer life insurance. Suppose 33%

offer both retirement plans and life insurance as benefits. If a small company is randomlyselected, determine the following probabilities:

a) The company offers a retirement plan given that they offer life insurance.b) The company offers a life insurance given that they offer retirement plan.c) The company offers a life insurance or a retirement plan.d) The company offers a retirement plan and does not offer life insurance.e) The company does not offer life insurance if it is known that they offer a retirement plan.

Let R = retirementLet L = life insurance

P(R) = .42 P(L) = .61 P(RC L) = .33

a) P(RL) = P(RC L)/P(L) = .33/.61 = .5410b) P(LR) = P(RC L)/P(R) = .33/.42 = .7857c) P(L B R) = P(L) + P(R) - P(L C R) =

.61 + .42 - .33 = .70

d) P(RC NL) = P(R) - P(RC L) = .42 - .33 = .09

e) P(NLR) = P(NL C R)/P(R) = .09/.42 = .2143

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4.39

According to link resources, 16% of the U.S population is technology-driven. However, thesefigures vary by region. For example, in the west the figure is 20% and in the northeast the figure

is 17%. 21% of the U.S. population in general is in the west and 20% of the U.S. population is in

the northeast. Suppose an American is chosen randomly.

a) What is the probability that the person lives in the west and is a technology-drivenperson?

b) What is the probability that the person lives in the northeast and is a technology-drivenperson?

c) Suppose the chosen person is known to be technology driven. What is the probability thatthe person lives in the west?

d) Suppose the chosen person is known not to be technology driven. What is the probabilitythat the person lives in the northeast?

e) Suppose the chosen person is known to be technology driven. What is the probability thatthe person lives in neither the west nor the northeast?

P(T) = .16 P(TW) = .20 P(TNE) = .17P(W) = .21 P(NE) = .20

a) P(W C T) = P(W)P(TW) = (.21)(.20) = .042b) P(NE C T) = P(NE)P(TNE) = (.20)(.17) = .034c) P(WT) = P(W C T)/P(T) = (.042)/(.16) = .2625d) P(NENT) = P(NE C NT)/P(NT) = {P(NE)P(NTNE)}/P(NT)

but P(NTNE) = 1 - P(TNE) = 1 - .17 = .83 and

P(NT) = 1 - P(T) = 1 - .16 = .84

Therefore, P(NENT) = {P(NE)P(NTNE)}/P(NT) =

{(.20)(.83)}/(.84) = .1976

e) P(not W C not NET) = P(not W C not NE C T)/ P(T)but P(not W C not NE C T) =

.16 - P(W C T) - P(NE C T) = .16 - .042 - .034 = .084

P(not W C not NE C T)/ P(T) = (.084)/(.16) = .525

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4.40

In a certain city, 30% of the families have a master card, 20% have an American express card,and 25% have a visa card. 8% of the families have both a master card and American express

card. 12% have both a visa card and master card. 6% have a both an American express card and

a visa card.

a) What is the probability of selecting a family that has either a visa card or an Americanexpress card?

b) If a family has a master card, what is the probability that it has a visa card?c) If a family has a visa card, what is the probability that it has a master card?d) Is possession of a visa card independent of possession of a master card? Why or why not?e) Is possession of an American express card mutually exclusive of possession of a visa

card?

Let M = Mastercard A = American Express V = Visa

P(M) = .30 P(A) = .20 P(V) = .25

P(M C A) = .08 P(V C M) = .12 P(A C V) = .06

a) P(V B A) = P(V) + P(A) - P(V C A)

= .25 + .20 - .06 = .39

b) P(VM) = P(V C M)/P(M) = .12/.30 = .40c) P(MV) = P(V C M)/P(V) = .12/.25 = .48

d) P(V) = P(VM)??

.25 { .40

Possession of Visa is not independent ofpossession of Mastercard

e) American Express is not mutually exclusive of Visa

because P(A C V) { .0000

4.41

A few years ago, a survey commissioned by the world almanac and maturity news service

reported that 51% of the respondents did not believe the social security system will be secure in20 years. Of the respondents who were age 45 or older, 70% believed the system will be secure

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in 20 years. Of the people surveyed, 57% were under age 45. One respondent is selectedrandomly.

a) What is the probability that the person is age 45 or older?b) What is the probability that the person is younger than age 45 and believes that the social

security system will be secure in 20 years?c) Is the person selected believes the social security system will be secure in 20 years, whatis the probability that the person is 45 years old or older?

d) What is the probability that the person is younger than age 45 or believes the socialsecurity system will not be secure in 20 years?

Let S = believe SS secure

N = don't believe SS will be secure45 = 45 or more years old

P(N) = .51Therefore, P(S) = 1 - .51 = .49

P(S>45) = .70Therefore, P(N>45) = 1 - P(S>45) = 1 - .70 = .30

P(45) = 1 - P(45 C S) = P(>45)P(S>45) = (.57)(.70) = .301

P(45 C S) = .49 - .301 = .189c) P(>45S) = P(>45 C S)/P(S) = P(>45)P(S>45)/P(S) =

(.43)(.70)/.49 = .6143

d) (

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4.42

A telephone survey conducted by the Maritz marketing research company found that 43% of theAmericans expert to save more money next year than they saved last year. 45% of those

surveyed plan to reduce debt next year. Of those who expect to save more money next year, 81%

plan to reduce debt next year. And American is selected randomly.

a) What is the probability that this person expect to save more money next year and plans toreduce debt next year?

b) What is the probability that this person expect to save more money next year or plans toreduce debt next year?

c) What is the probability that this person neither expect to save more money next year norplans to reduce debt next year?

d) What is the probability that this person expect to save more money next year and does notplans to reduce debt next year?

Let M = expect to save more

R = expect to reduce debtNM = don't expect to save more

NR = don't expect to reduce debt

P(M) = .43 P(R) = .45 P(RM) = .81P(NRM) = 1-P(RM) = 1 - .81 = .19P(NM) = 1 - P(M) = 1 - .43 = .57

P(NR) = 1 - P(R) = 1 - .45 = .55

a) P(M C R) = P(M)P(RM) = (.43)(.81) = .3483b) P(M B R) = P(M) + P(R) - P(M C R)

= .43 + .45 - .3483 = .5317

c) P(neither save nor reduce debt) =

1 - P(M B R) = 1 - .5317 = .4683

d) P(M C NR) = P(M)P(NRM) = (.43)(.19) = .0817

Probability matrix for problem 4.42:

ReduceYes No

Save

Yes .3483 .0817 .43

No .1017 .4683 .57

.4500 .5500 1.00

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4.43

The steel case workplace index studied the type of work related activities that Americans did

while on vacation of the summer. Among other things, 40% read work related material. 34%checked in with the boss. Respondent to the study were allowed to select more than one activity.

Suppose that of those who read work related material, 78% checked in with the boss. One ofthese survey respondents is selected randomly.

a) What is the probability that while on vacation this respondent checked in with the bossand read work related material?

b)

What is the probability that while on vacation this respondent neither read work-relatedmaterial nor checked in with the boss?

c) What is the probability that while on vacation this respondent read work-related materialgiven that the respondent checked in with the boss?

d) What is the probability that while on vacation this respondent did not check in with theboss given that the respondent read work-related material?

e) What is the probability that while on vacation this respondent did not check in with theboss given that the respondent did not read work-related material?

f) Construct a probability matrix for this problem.

Let R = readLet B = checked in the with boss

P(R) = .40 P(B) = .34 P(BR) = .78

a) P(B C R) = P(R)P(BR) = (.40)(.78) = .312b) P(NRC NB) = 1 - P(RB B)

but P(RB B) = P(R) + P(B) - P(RC B) =

.40 +.34 - .312 = .428

P(NRC NB) = 1 - .428 = .572c) P(RB) = P(RC B)/P(B) = (.312)/(.34) = .9176

d) P(NBR) = 1 - P(BR) = 1 - .78 = .22e) P(NBNR) = P(NB C NR)/P(NR)

but P(NR) = 1 - P(R) = 1 - .40 = .60

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P(NBNR) = .572/.60 = .9533

f) Probability matrix for problem 4.43:

B NBR .312 .088 .40

NR .028 .572 .60

.340 .660 1.00

4.44

Health rights hotline published the result of a survey of 2400 people in northern California inwhich consumers were asked to share their complaints about managed care. The number one

complaint was denial of care, with 17% of the participating consumers selecting it. Several othercomplaints were noted including inappropriate care (14%), consumers service (14%), payment

disputes (11%), specialty care (10%), delays in getting care (8%), and prescription drugs (7%).These complaints categories are mutually exclusive. Assume that the result of this survey can be

inferred to all managed care consumers. If a managed care consumer is randomly selected,determine the following probabilities:

a) The consumer complaints about payment disputes or specialty care.b) The consumer complaints about prescription drugs and consumer service.c) The consumer complaints about inappropriate care given that the consumer complaints

about specialty care.d) The consumer does not complain about delays in getting care nor does the consumer

complain about payment disputes.

Let: D = denial

I = inappropriateC = customer

P = payment disputeS = specialty

G = delays getting careR = prescription drugs

P(D) = .17 P(I) = .14 P(C) = .14 P(P) = .11

P(S) = .10 P(G) = .08 P(R) = .07

a) P(P B S) = P(P) + P(S) = .11 + .10 = .21

b) P(RC C) = .0000 (mutually exclusive)

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c) P(IS) = P(I C S)/P(S) = .0000/.10 = .0000d) P(NG C NP) = 1 - P(G B P) = 1 - [P(G) + P(P)] =

1 [.08 + .11] = 1 - .19 = .81

4.45

Companies use employee training for various reasons including employee loyalty, certification,

quality, and process improvement. In a national survey of companies, B1 learning Systemsreported that 56% percent of the responding companies named employee retention as a top

reason for training suppose 36% of the companies replied that they use training for processimprovement and for employee retention. In addition, suppose that of the companies that use

training for process improvement 90% use training for employee retention. A company that usestraining is randomly selected.

a) What is the probability that the company uses training for employee retention and not forprocess improvement?

b) If it is known that the company uses training for employee retention, what is theprobability that it uses training for process improvement?

c) What is the probability that the company uses training for process improvement?d) What is the probability that the company uses training for employee retention or process

improvement?e) What is the probability that the company neither uses training for employee retention nor

uses training for process improvement?f) Suppose it is known that the company does not use training for process improvement.

What is the probability that the company dose use training for employee retention?

Let R = retention

P = process

P(R) = .56 P(P C R) = .36 P(RP) = .90

a) P(RC NP) = P(R) - P(P C R) = .56 - .36 = .20

b) P(PR) = P(P C R)/P(R) = .36/.56 = .6429

c) P(P) = ??

P(RP) = P(RC P)/P(P)so P(P) = P(RC P)/P(RP) = .36/.90 = .40d) P(RB P) = P(R) + P(P) - P(RC P) =

.56 + .40 - .36 = .60

e) P(NRC NP) = 1 - P(RB P) = 1 - .60 = .40

f) P(RNP) = P(RC NP)/P(NP)

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but P(NP) = 1 - P(P) = 1 - .40 = .60

P(RNP) = .20/.60 = .33

4.46

Pitney Bowes surveyed 302 directors and vice presidents of marketing at large and midsized U.S.companies to determine what they believe is the best vehicle for educating decision makers on

complex issues in selling products and services. The highest percentage of companies chosedirect mail/catalogs, followed by direct sales/sales rep. direct mail/catalogs was selected by 38%

of the companies. None of the companies selected both direct mail/catalogs and direct sales/salesrep. Suppose also that 41% selected neither direct mail/catalogs nor direct sales/sales rep. if one

of these companies is selected randomly and their top marketing person interviewed about thismatter, determine the following probabilities:

a) The marketing person selected direct mail/catalogs and did not select direct sales/salesrep.

b) The marketing person selected direct sales/sales rep.c) The marketing person selected direct sales/sales rep. given that the person selected direct

mail/catalogs.

d)

The marketing person did not select direct mail/catalogs given that the person did notselect direct sales/sales rep.

Let M = mail

S = sales

P(M) = .38 P(M C S) = .0000 P(NM C NS) = .41

a) P(M C NS) = P(M) - P(M C S) = .38 - .00 = .38

b) P(S):

P(M B S) = 1 - P(NM C NS) = 1 - .41 = .59P(M B S) = P(M) + P(S)

Therefore, P(S) = P(M B S) - P(M) = .59 - .38 = .21

c) P(SM) = P(S C M)/P(M) = .00/.38 = .00

d) P(NMNS) = P(NM C NS)/P(NS) = .41/.79 = .5190

where: P(NS) = 1 - P(S) = 1 - .21 = .79

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4.47

A small independent physicians practice has three doctors. Dr.Sarabia sees 41% of the patients,Dr. tran sees 32% and Dr.jackson sees the rest. Dr.sarabia requests blood tests on 5% of her

patients, Dr.tren requests blood tests on 8% of his patients, and Dr. Jackson requests blood testson 6% of her patients. An auditor randomly selects a patient from the past week and discovers

that the patient had a blood test as a result of the physician visit. Knowing this information, whatis the probability that the patient saw Dr. Sarabia? For what percentage of all patients at this

practice are blood tests requested?

Let S = SarabiaT = Tran

J = JacksonB = blood test

P(S) = .41 P(T) = .32 P(J) = .27

P(B S) = .05 P(B T) = .08 P(B J) = .06

Event Prior Conditional Joint Revised

P(Ei) P(BEi)P(B C Ei)

P(BiNS)S .41 .05 .0205 .329

T .32 .08 .0256 .411

J .27 .06 .0162 .260P(B)=.0623

4.48

A survey by the Arthur Andersen enterprise group/national small business united attempted todetermine what the leading challenges are for the growth and survival of small businesses.

Although the economy and finding qualified workers were the leading challenges, several otherwere listed in the results of the study, including regulations, listed by 30% of the companies, and

the tax burden, listed by 35%. Suppose that 71% of the companies listing regulations as achallenge listed the tax burden as a challenge. Assume these percentages hold for all small

businesses. If a small business is randomly selected, determine the following probabilities:

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a) The small business lists both the tax burden and regulations as a challenge.b) The small business lists either the tax burden or regulations as a challenge.c) The small business lists either the tax burden or regulations but not both as a challenge.d)

The small business lists regulations as a challenge given that it lists the tax burden as achallenge.

e) The small business dose not list regulations as a challenge given that it lists the taxburden as a challenge.

f) The small business does not list regulations as a challenge given that it does not list thetax burden as a challenge.

Let R = regulations

T = tax burden

P(R) = .30 P(T) = .35 P(TR) = .71

a) P(RC T) = P(R)P(TR) = (.30)(.71) = .2130b) P(RB T) = P(R) + P(T) - P(RC T) =

.30 + .35 - .2130 = .4370

c) P(RB T) - P(RC T) = .4370 - .2130 = .2240

d) P(RT) = P(RC T)/P(T) = .2130/.35 = .6086

e) P(NRT) = 1 - P(RT) = 1 - .6086 = .3914f) P(NRNT) = P(NRC NT)/P(NT) = [1 - P(RB T)]/P(NT) =

(1 - .4370)/.65 = .8662

4.49

According to the public for food and health policy, approximately 27% of all soup products in arecent year did not carry nutritional labeling. Approximately 83% of hot dog products did nothave nutritional labeling. Assume that if there groups of foods were combined, 60% would be

soup products, 35% would be breakfast meats, and 5% would be hot dogs. A researcher isblindly given a food product from one of these three groups and is told that the product does

have nutritional labeling. Revise the probabilities that the product is a soup product, a breakfastmeat, and a hot dog product.

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Event Prior Conditional Joint Revised

P(Ei) P(NSEi)P(NS C Ei)

P(EiNS)Soup .60 .73 .4380 .8456

BreakfastMeats .35 .17 .0595 .1149

Hot Dogs .05 .41 .0205 .0396

.5180

4.50

A survey conducted for lifetimes daily half-hour series the great American TV poll askedAmerican what they consider to be the most important thing in their lives. Twenty-nine percent

said good health, 21% responded a happy marriage, and 40% replied faith in god. Becausethey were asked which of these things the most important thing is, a respondent could not select

more than one answer.

a) What is the probability that a person replied a happy marriage or faith in god?b) What is the probability that a person replied a happy marriage or faith in god or good

health?

c) What is the probability that a person replied a happy marriage and faith in god?d) What is the probability that a person replied neither a happy marriage nor faith ingod nor good health?

Let GH = Good health

HM = Happy marriageFG = Faith in God

P(GH) = .29 P(HM) = .21 P(FG) = .40

a) P(HM B FG) = P(HM) + P(FG) - P(HM C FG)but P(HM C FG) = .0000

P(HM B FG) = P(HM) + P(FG) = .21 + .40 = .61

b) P(HM B FG B GH) = P(HM) + P(FG) + P(GH) =

.29 + .21 + .40 = .9000

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c) P(FG C GH) = .0000

The categories are mutually exclusive.

The respondent could not select more than oneanswer.

d) P(neither FG nor GH nor HM) =

1 - P(HM B FG B GH) = 1 - .9000 = .1000