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Obtain an understanding of the role probability information plays in the decision making process.

2.

Understand probability as a numerical measure of the likelihood of occurrence.

3.

Be able to use the three methods (classical, relative frequency, and subjective) commonly used for assigning probabilities and understand when they should be used.

4.

Be able to use the addition law and be able to compute the probabilities of events using conditional probability and the multiplication law.

5.

Be able to use new information to revise initial (prior) probability estimates using Bayes' theorem.

6.

Know the definition of the following terms: experiment sample space event complement Venn Diagram union of events intersection of events Bayes' theorem

addition law mutually exclusive conditional probability independent events multiplication law prior probability posterior probability

2-1

Chapter 2

Solutions: 1.

a.

Go to the x-ray department at 9:00 a.m. and record the number of persons waiting.

b.

The experimental outcomes (sample points) are the number of people waiting: 0, 1, 2, 3, and 4. Note: While it is theoretically possible for more than 4 people to be waiting, we use what has actually been observed to define the experimental outcomes.

c. Number Waiting 0 1 2 3 4 Total:

2.

Probability .10 .25 .30 .20 .15 1.00

d.

The relative frequency method was used.

a.

Choose a person at random, have them taste the 4 blends and state their preference.

b.

Assign a probability of 1/4 to each blend. We use the classical method of equally likely outcomes here.

c. Blend 1 2 3 4 Total:

Probability .20 .30 .35 .15 1.00

The relative frequency method was used. 3.

4.

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. a.

Use the relative frequency approach: P(California) = 1,434/2,374 = .60

b.

Number not from 4 states = 2,374 - 1,434 - 390 - 217 - 112 = 221 P(Not from 4 States) = 221/2,374 = .09

c.

P(Not in Early Stages) = 1 - .22 = .78

d.

Estimate of number of Massachusetts companies in early stage of development - (.22)390 ≈ 86

2-2

Introduction to Probability

e.

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 billion Authors' Note: The actual amount going to Colorado was $1.74 billion.

5.

6.

7.

a.

No, the probabilities do not sum to one. They sum to 0.85.

b.

Owner must revise the probabilities so that they sum to 1.00.

a.

P(A) = P(150 - 199) + P(200 and over) 26 5 + = 100 100 = 0.31

b.

P(B) = P(less than 50) + P(50 - 99) + P(100 - 149) = 0.13 + 0.22 + 0.34 = 0.69

a.

P(A) = .40, P(B) = .40, P(C) = .60

b.

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) = .40

d.

A ∪ Bc = {E1, E2, E5} P(A ∪ Bc) = .60

e.

P(B ∪ C) = P(E2, E3, E4, E5) = .80

8.

Let Y = high one-year return M = high five-year return a.

P(Y) = 15/30 = .50 P(M) = 12/30 = .40 P(Y ∩ M) = 6/30 = .20

b.

P(Y ∪ M) = P(Y) + P(M) - P(Y ∩ M) = .50 + .40 - .20 = .70

c.

1 - P(Y ∪ M) = 1 - .70 = .30

2-3

Chapter 2

9.

Let E = event patient treated experienced eye relief. S = event patient treated had skin rash clear up. Given: P (E)

= 90 / 250 = 0.36

P (S)

= 135 / 250 = 0.54

P (E ∪ S)

= 45 / 250 = 0.18

P (E ∪ S ) = P (E) + P (S) - P (E ∩ S) = 0.36 + 0.54 - 0.18 = 0.72 10.

Let J D

= event student had a part-time job. = event student made the Dean’s list.

Given: P (J)

= 40 / 100 = 0.40

P (D)

= 25 / 100 = 0.25

P (J ∩ D)

= 15 / 100 = 0.15

P (J ∪ D ) = P (J) + P (D) - P (J ∩ D) = 0.40 + 0.25 - 0.15 = 0.50 11. a. b.

Yes; the person cannot be in an automobile and a bus at the same time. P(Bc) = 1 - P(B) = 1 - 0.35 = 0.65

12. a.

P(A B) =

P(A ∩ B) 0.40 = = 0.6667 P(B) 0.60

b.

P(B A) =

P(A ∩ B) 0.40 = = 0.80 P(A) 0.50

c.

No because P(A | B) ≠ P(A)

13. a.

Full Time Part Time Total

Quality 0.218 0.208 0.426

Reason for Applying Cost/Convenience 0.204 0.307 0.511

Other 0.039 0.024 0.063

Total 0.461 0.539 1.00

b.

It is most likely a student will cite cost or convenience as the first reason: probability = 0.511. School quality is the first reason cited by the second largest number of students: probability = 0.426.

c.

P (Quality⏐full time) = 0.218/0.461 = 0.473

2-4

Introduction to Probability

d.

P (Quality⏐part time) = 0.208/0.539 = 0.386

e.

P (B) = 0.426 and P (B⏐A) = 0.473 Since P (B) ≠ P (B⏐A), the events are dependent.

14. a.

P(O) = 0.38 + 0.06 = 0.44

b.

P(Rh-) = 0.06 + 0.02 + 0.01 + 0.06 = 0.15

c.

P(both Rh-) = P(Rh-) P(Rh-) = (0.15)(0.15) = 0.0225

d.

P(both AB) = P(AB) P(AB) = (0.05)(0.05) = 0.0025

e.

P(Rh- O) =

f.

P(Rh+) = 1 - P(Rh-) = 1 - 0.15 = 0.85 P(B Rh+) =

15. a.

P(Rh- ∩ O) 0.06 = = 0.136 P(O) 0.44

P(B ∩ Rh+) 0.09 = = 0.106 P(Rh+) 0.85

Total sample size = 2000 Dividing each entry by 2000 provides the following joint probability table.

Age 18 to 34 35 and over

Let

b.

A= B= Y= N=

Health Insurance Yes No Total .375 .085 .46 .475 .065 .54 .850 .150 1.00

18 to 34 age group 35 and over age group Insurance coverage No insurance coverage

P(A) = .46 P(B) = .54 Of population age 18 and over 46% are ages 18 to 34 54% are ages 35 and over

c.

P(N) = .15

d.

P(N A) =

P(N ∩ A) .085 = = .1848 P(A) .46

e.

P(N B) =

P(N ∩ B) .065 = = .1204 P(B) .54

2-5

Chapter 2 P(A ∩ N) .085 = = .5677 P(N) .150

f.

P(A N) =

g.

Probability of no health insurance coverage is .15. A higher probability exists for the younger population. Ages 18 to 34: .1848 or approximately 18.5% of the age group. Ages 35 and over: .1204 or approximately 12% of the age group. Of the no insurance group, more are in the 18 to 34 age group: .5677, or approximately 57% are ages 18 to 34.

16. a.

P(A ∩ B) = P(A)P(B) = (0.55)(0.35) = 0.19

b.

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.90 - 0.19 = 0.71

c.

1 - 0.71 = 0.29

17. a. Occupation Cabinetmaker Lawyer Physical Therapist Systems Analyst

Under 50 .000 .150 .000 .050 Total .200

50-59 .050 .050 .125 .025 .250

Satisfaction Score 60-69 70-79 .100 .075 .025 .025 .050 .025 .100 .075 .275 .200

b.

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)

18. a.

P(B) = 0.25 P(S⏐B) = 0.40 P(S ∩ B) = 0.25(0.40) = 0.10 P(S ∩ B) 0.10 = = 0.25 P(S) 0.40

b.

P(B S) =

c.

B and S are independent. The program appears to have no effect.

19.

Let:

A = lost time accident in current year B = lost time accident previous year

∴ Given: P(B) = 0.06, P(A) = 0.05, P(A⏐B) = 0.15

a.

P(A ∩ B) = P(A⏐B)P(B) = 0.15(0.06) = 0.009

2-6

80-89 .025 .000 .050 .000 .075

Total .250 .250 .250 .250 1.000

Introduction to Probability

b.

20. a.

P(A ∪ B) = P(A) + P(B) - P(A⏐ B) = 0.06 + 0.05 - 0.009 = 0.101 or 10.1% P(B ∩ A1) = P(A1)P(B⏐A1) = (0.20)(0.50) = 0.10 P(B ∩ A2) = P(A2)P(B⏐A2) = (0.50)(0.40) = 0.20 P(B ∩ A3) = P(A3)P(B⏐A3) = (0.30)(0.30) = 0.09

b.

P(A 2 B) =

0.20 = 0.51 0.10 + 0.20 + 0.09

c.

21.

Events

P(Ai)

P(B⏐Ai)

P(Ai ∩ B)

P(Ai⏐ B)

A1 A2 A3

0.20 0.50 0.30 1.00

0.50 0.40 0.30

0.10 0.20 0.09 0.39

0.26 0.51 0.23 1.00

S1 = successful, S2 = not successful and B = request received for additional information. a.

P(S1) = 0.50

b.

P(B⏐S1) = 0.75

c.

P(S1 B) =

22. a. b.

(0.50)(0.75) 0.375 = = 0.65 (0.50)(0.75) + (0.50)(0.40) 0.575

Let F = female. Using past history as a guide, P(F) = .40 Let D = Dillard's P(F D) =

.40(3 / 4) .30 = = .67 .40(3 / 4) + .60(1/ 4) .30 + .15

The revised (posterior) probability that the visitor is female is .67. We should display the offer that appeals to female visitors. 23. a. b.

P(Oil) = 0.50 + 0.20 = 0.70 Let S = Soil test results Events

P(Ai)

P(S⏐Ai)

High Quality (A1) Medium Quality (A2) No Oil (A3)

0.50 0.20 0.30 1.00

0.20 0.80 0.20

2-7

P(Ai ∩ S) 0.10 0.16 0.06 P(S) = 0.32

P(Ai⏐ S) 0.31 0.50 0.19 1.00

Chapter 2

P(Oil) = 0.81 which is good; however, probabilities now favor medium quality rather than high quality oil. 24.

Let: S = small car Sc = other type of vehicle F = accident leads to fatality for vehicle occupant We 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: Prior Probabilities .18 .82 1.00

Events S Sc

Conditional Probabilities .128 .050

Joint Probabilities .023 .041 .064

Posterior Probabilities .36 .64 1.00

From 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. 25. Events Supplier A Supplier B Supplier C

P(Ai)

P(D⏐Ai)

0.60 0.30 0.10 1.00

0.0025 0.0100 0.0200

a.

P(D) = 0.0065

b.

B is the most likely supplier if a defect is found.

P(Ai∩D) 0.0015 0.0030 0.0020 P(D) = 0.0065

26. a. Events

P(Di)

P(S1|Di)

D1 D2

.60 .40 1.00

.15 .80

P(Di ∩S1) .090 .320 P(S1) = .410

P(Di |S1) .2195 .7805 1.0000

P(D1 | S1) = .2195 P(D2 | S1) = .7805 b. Events

P(Di)

P(S2 |Di)

D1 D2

.60 .40 1.00

.10 .15

P(D1 | S2) = .50 P(D2 | S2) = .50

2-8

P(Di ∩S2) .060 .060 P(S2) = .120

P(Di |S2) .500 .500 1.000

P(Ai⏐D) 0.23 0.46 0.31 1.00

Introduction to Probability

c. Events

P(Di)

P(S3 |Di)

D1 D2

.60 .40 1.00

.15 .03

P(Di ∩S3) .090 .012 P(S3) = .102

P(Di |S3) .8824 .1176 1.0000

P(D1 | S3) = .8824 P(D2 | S3) = .1176 d.

Use the posterior probabilities from part (a) as the prior probabilities here. Events

P(Di)

P(S2 | Di)

P(Di ∩ S2)

P(Di | S2)

D1 D2

.2195 .7805 1.0000

.10 .15

.0220 .1171 .1391

.1582 .8418 1.0000

P(D1 | S1 and S2) = .1582 P(D2 | S1 and S2) = .8418

2-9

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