The Rules of Probability Measuring Probabilities

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4-22. Several years ago, a large state university conducted a survey of undergraduate students regarding their use of computers. The results of the survey are contained in the data file ComputerUse.

a. Based on the data from the survey, what was the probability that undergraduate students at this university will have a major that requires them to use a computer on a daily basis?

b. Based on the data from this survey, if a student was a business major, what was the probability of the student believing that the computer lab facilities are very adequate?

c. Select a sample of 20 students at your university and ask them what their major is and whether the major requires them to access the Internet for computational purposes, with a personal computer, notebook, tablet, or other device. Use these data to calculate the probability of each major requiring a computing device, and then compare your results with those from the older survey.

4-23. A scooter manufacturing company is notified whenever a scooter breaks down, and the problem is classified as being either mechanical or electrical. The company then matches the scooter to the plant where it was assembled. The file Scooters contains a random sample of 200 breakdowns. Use the data in the file and the relative frequency assessment method to find the following probabilities:

a. What is the probability a scooter was assembled at the Tyler plant?

b. What is the probability that a scooter breakdown was due to a mechanical problem?

c. What is the probability that a scooter was assembled at the Lincoln plant and had an electrical problem?

4-24. A survey on cell phone use asked, in part, what was the most important reason people give for not using a wireless phone exclusively. The responses were:

(1) Like the safety of traditional phone, (2) Like having a land line for a backup phone, (3) Pricing not

attractive enough, (4) Weak or unreliable cell signal at home, (5) Need phone line for DSL Internet access, and (6) Other. The file titled Wireless contains the responses for the 1,088 respondents.

a. Calculate the probability that a randomly chosen respondent would not use a wireless phone exclusively because of some type of difficulty in placing and receiving calls with a wireless phone.

b. Calculate the probability that a randomly chosen person would not use a wireless phone exclusively because of some type of difficulty in placing and receiving calls with a wireless phone and is over the age of 55.

c. Determine the probability that a randomly chosen person would not use a wireless phone exclusively because of a perceived need for Internet access and the safety of a traditional phone.

d. Of those respondents under 36, determine the probability that an individual in this age group would not use a wireless phone exclusively because of some type of difficulty in placing and receiving calls with a wireless phone.

4-25. A selection of soft-drink users is asked to taste two disguised soft drinks and indicate which they prefer.

The file titled Challenge contains the results of a study that was conducted on a college campus.

a. Determine the probability that a randomly chosen student prefers Pepsi.

b. Determine the probability that one of the students prefers Pepsi and is less than 20 years old.

c. Of those students who are less than 20 years old, calculate the probability that a randomly chosen student prefers (1) Pepsi and (2) Coke.

d. Of those students who are at least 20 years old, calculate the probability that a randomly chosen student prefers (1) Pepsi and (2) Coke.

The Rules of Probability

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All possible outcomes associated with an experiment form the sample space. Therefore, the sum of the probabilities of all possible outcomes is 1, as shown by Probability Rule 2.

Addition Rule for Individual Outcomes If a single event is made up of two or more individual outcomes, then the probability of the event is found by summing the probabilities of the individual outcomes. This is illustrated by Probability Rule 3.

Probability Rule 2

a

k i=1

P1e_i2 = 1 where:

k = Number of outcomes in the sample e_i = ith outcome

(4.4)

Probability Rule 3: Addition Rule for Individual Outcomes

The probability of an event E_i is equal to the sum of the probabilities of the individual outcomes that form E_i. For example, if

E_i = 5e₁, e₂, e₃6 then

P1E_i2 = P1e₁2 + P1e₂2 + P1e₃2 (4.5)

TABLE 4.2 Google’s Survey Results

Searches per Day Frequency Relative Frequency

e₁ =At least 10 400 0.08

e₂ = 3 to 9 1,900 0.38

e₃ = 1 to 2 1,500 0.30

e₄ = 0 1,200 0.24

Total 5,000 1.00

BUSINESS APPLICATION

Addition Rule

Google Google has become synonymous with Web searches and is the leader in the search engine marketplace. Suppose officials at the northern California headquarters have recently performed a survey of computer users to determine how many Internet searches individuals do daily using Google. Table 4.2 shows the results of the survey of Internet users.

The sample space for the experiment for each respondent is SS = 5e₁, e₂, e₃, e₄6 where the possible outcomes are

e₁ = At least 10 searches e₂ = 3 to 9 searches e₃ = 1 to 2 searches e₄ = 0 searches Probability Rule 1

For any event E_i,

0 … P1E_i2 … 1 for all i (4.3)

o u tc o m e 2

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EXAMPLE 4-7

The Addition Rule for Individual Outcomes

KQRT 1340 Radio The KQRT 1340 radio station is a combination news/talk and “oldies” station. During a 24-hour day, a listener can tune in and hear any of the following four programs being broadcast:

“Oldies” music News stories Talk programming Commercials

Recently, the station has been having trouble with its transmitter. Each day, the sta- tion’s signal goes dead for a few seconds; it seems that these outages are equally likely to occur at any time during the 24-hour broadcast day. There seems to be no pattern regard- ing what is playing at the time the transmitter problem occurs. The station manager is concerned about the probability that these problems will occur during either a news story or a talk program.

s t e p 1 Define the experiment.

The station conducts its broadcast starting at 12:00 midnight, extending until a transmitter outage is observed.

s t e p 2 Define the possible outcomes.

The possible outcomes are each type of programming that is playing when the transmitter outage occurs. There are four possible outcomes:

e₁ = Oldies e₂ = News

e₃ = Talk programs e₄ = Commercials

s t e p 3 Determine the probability of each possible outcome.

The station manager has determined that out of the 1,440 minutes per day, 540 minutes are oldies, 240 minutes are news, 540 minutes are talk programs, and 120 minutes are commercials. Therefore, the probability of each type of programming being on at the moment the outage occurs is assessed as follows:

Using the relative frequency assessment approach, we assign the following probabilities:

P1e₁2 = 400>5,000 = 0.08 P1e₂2 = 1,900>5,000 = 0.38 P1e₃2 = 1,500>5,000 = 0.30 P1e₄2 = 1,200>5,000 = 0.24 g = 1.00

Assume we are interested in the event respondent performs 1 to 9 searches per day:

E = Internet user performs 1 to 9 searches per day The outcomes that make up E are

E = 5e₂, e₃6

We can find the probability, P(E), by using Probability Rule 3 (Equation 4.5), as follows:

P1E 2 = P1e₂2 + P1e₃2

= 0.38 + 0.30

= 0.68

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Outcome P1e_i2

e₁ = Oldies

P(e₁) = 540

1,440 = 0.375 e₂ = News

P(e₂) = 240

1,440 = 0.167 e₃ = Talk programs

P(e₃) = 540

1,440 = 0.375 e₄ = Commercials

P(e₄) = 120

1,440 = 0.083 a = 1.000

Note that based on Equation 4.4 (Probability Rule 2), the sum of the probabilities of the individual possible outcomes is 1.0.

s t e p 4 Define the event of interest.

The event of interest is a transmitter problem occurring during a news or talk program. This is

E = 5e₂, e₃6

s t e p 5 Use Probability Rule 3 (Equation 4.5) to compute the desired probability.

P1E 2 = P1e₂2 + P1e₃2 P1E 2 = 0.167 + 0.375 P1E 2 = 0.542

Thus, the probability is slightly higher than 0.5 that when a transmitter problem occurs, it will happen during either a news or talk program.

TRY EXERCISE 4-26 (pg. 184)

Complement Rule Closely connected with Probability Rules 1 and 2 is the complement of an event. The complement of event E is represented by E. The Complement Rule is a corollary to Probability Rules 1 and 2.

Complement

The complement of an event E is the collection of all possible outcomes not contained in event E.

Complement Rule

P1E 2 = 1 - P1E 2

That is, the probability of the complement of event E is 1 minus the probability of event E.

(4.6)

EXAMPLE 4-8

The Complement Rule

Capital Consulting The managing partner for Capital Consulting is working on a proposal for a consulting project with a client in Sydney, Australia. The manager lists four possible net profits from the consulting engagement and his subjectively assessed proba- bilities related to each profit level.

Outcome P(Outcome)

$ 0 0.70

$ 2,000 0.20

$15,000 0.07

$50,000 0.03

g = 1.00

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Note that each probability is between 0 and 1 and that the sum of the probabilities is 1, as required by Probability Rules 1 and 2.

The manager plans to submit the proposal if the consulting engagement will have a positive profit, so he is interested in knowing the probability of an outcome greater than

$0. This probability can be found using the Complement Rule with the following steps:

s t e p 1 Determine the probabilities for the outcomes.

P1+02 = 0.70 P1+2,0002 = 0.20 P1+15,0002 = 0.07 P1+50,0002 = 0.03

s t e p 2 Find the desired probability.

Let E be the consulting outcome event =+0. The probability of the $0 outcome is

P1E2 = 0.70

The complement, E, is all investment outcomes greater than $0. From the Complement Rule, the probability of profit greater than $0 is

P1Profit 7 +02 = 1 - P1Profit = +02 P1Profit 7 +02 = 1 - 0.70

P1Profit 7 +02 = 0.30

Based on the manager’s subjective probability assessment, there is a 30%

chance the consulting project will have a positive profit.

TRY EXERCISE 4-32 (pg. 184)

Addition Rule for Any Two Events

BUSINESS APPLICATION

Addition Rule

Google (continued ) Suppose the staff who conducted the survey for Google discussed ear- lier also asked questions about the computer users’ ages. The Google managers consider age important in designing their search engine methodologies. Table 4.3 shows the breakdown of the sample by age group and by the number of times a user performs a search each day.

Table 4.3 shows that seven events are defined. For instance, E₁ is the event that a com- puter user performs ten or more searches per day. This event is composed of three individual outcomes associated with the three age categories. These are

E₁ = 5e₁, e₂, e₃6

In another case, event E₅ corresponds to a survey respondent being younger than 30 years of age. It is composed of four individual outcomes associated with the four levels of search activity. These are

E₅ = 5e₁, e₄, e₇, e₁₀6

o u tc o m e 2

TABLE 4.3 Google Search Study

Age Group

Searches per Day E₅

Under 30 E₆

30 to 50 E₇

Over 50 Total E₁ Ú 10 searches e₁ 200 e₂ 100 e₃ 100 400 E₂ 3 to 9 searches e₄ 600 e₅ 900 e₆ 400 1,900 E₃ 1 to 2 searches e₇ 400 e₈ 600 e₉ 500 1,500 E₄ 0 searches e₁₀ 700 e₁₁ 500 e₁₂ 0 1,200

Total 1,900 2,100 1,000 5,000

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Suppose we wish to find the probability of E₄ (0 searches) or E₆ (being in the 30-to-50 age group). That is,

P1E₄ or E₆2 = ?

TABLE 4.4 Google—Joint Probability Table

Age Group

Searches per Day E ₅ Under 30 E ₆ 30 to 50 E ₇ Over 50 Total

E₁ Ú 10 searches e₁ 200>5,000 = 0.04 e₂ 100>5,000 = 0.02 e₃ 100>5,000 = 0.02 400>5,000 = 0.08 E₂ 3 to 9 searches e₄ 600>5,000 = 0.12 e₅ 900>5,000 = 0.18 e₆ 400>5,000 = 0.08 1,900>5,000 = 0.38 E₃ 1 to 2 searches e₇ 400>5,000 = 0.08 e₈ 600>5,000 = 0.12 e₉ 500>5,000 = 0.10 1,500>5,000 = 0.30 E₄ 0 searches e₁₀ 700>5,000 = 0.14 e₁₁ 500>5,000 = 0.10 e₁₂ 0>5,000 = 0.00 1,200>5,000 = 0.24 Total 1,900>5,000 = 0.38 2,100>5,000 = 0.42 1,000>5,000 = 0.20 5,000>5,000 =1.00

To find this probability, we must use Probability Rule 4.

The key word in knowing when to use Rule 4 is or. The word or indicates addition. Figure 4.1 is a Venn diagram that illustrates the application of the Addition Rule for Any Two Events.

Notice that the probabilities of the outcomes in the overlap between the two events, E₁ and E₂, are double-counted when the probabilities of the outcomes in E₁ are added to those in E₂. Thus, the probabilities of the outcomes in the overlap, which is E₁ and E₂, need to be sub- tracted to avoid the double-counting.

Probability Rule 4: Addition Rule for Any Two Events E₁ and E₂

P1E₁ or E₂2 = P1E₁2 + P1E₂2 - P1E₁ and E₂2 (4.7)

E1 E₂

E1and E2

P(E1or E2) = P(E1) + P(E²) – P(E1and E2) FIGURE 4.1 Venn Diagram—

Addition Rule for Any Two Events

Table 4.3 illustrates two important concepts in data analysis: joint frequencies and mar- ginal frequencies. Joint frequencies, which were discussed in Chapter 2, are the values inside the table. They provide information on age group and search activity jointly. Marginal fre- quencies are the row and column totals. These values give information on only the age group or only Google search activity.

For example, 2,100 people in the survey are in the 30- to 50-year age group. This column total is a marginal frequency for the age group 30 to 50 years, which is represented by E₆. Now notice that 600 respondents are younger than 30 years old and perform three to nine searches a day. The 600 is a joint frequency whose outcome is represented by e₄.

Table 4.4 shows the relative frequencies for the data in Table 4.3. These values are the probability assessments for the events and outcomes.

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Referring to the Google situation, the probability of E₄ (0 searches) or E₆ (being in the 30-to-50 age group) is

P1E₄ or E₆2 = ?

Table 4.5 shows the relative frequencies with the events of interest shaded. The overlap cor- responds to the joint occurrence (intersection) of conducting 0 searches and being in the 30-to-50 age group. The probability of the outcomes in the overlap is represented by P1E₄ and E₆2 and must be subtracted. This is done to avoid double-counting the probabilities of the outcomes that are in both E₄ and E₆ when calculating the P1E₄ or E₆2. Thus,

P1E₄ or E₆2 = P1E₄2 + P1E₆2 - P1E₄ and E₆2

= 0.24 + 0.42 - 0.10

= 0.56

Therefore, the probability that a respondent will either be in the 30-to-50 age group or per- form 0 searches on a given day is 0.56.

What is the probability a respondent will perform 1 to 2 searches or be in the over-50 age group? Again, we can use Probability Rule 4:

P1E₃ or E₇2 = P1E₃2 + P1E₇2 - P1E₃ and E₇2 Table 4.6 shows the relative frequencies for these events. We have

P1E₃ or E₇2 = 0.30 + 0.20 - 0.10 = 0.40

Thus, there is a 0.40 chance that a respondent will perform 1 to 2 searches or be in the over- 50 age group.

TABLE 4.5 Google Searches—Addition Rule Example

Age Group

Searches per Day E₅ Under 30 E₆ 30 to 50 E₇ Over 50 Total

E₁ Ú 10 searches e₁ 200>5,000 = 0.04 e₂ 100>5,000 = 0.02 e₃ 100>5,000 = 0.02 400>5,000 = 0.08 E₂ 3 to 9 searches e₄ 600>5,000 = 0.12 e₅ 900>5,000 = 0.18 e₆ 400>5,000 = 0.08 1,900>5,000 = 0.38 E₃ 1 to 2 searches e₇ 400>5,000 = 0.08 e₈ 600>5,000 = 0.12 e₉ 500>5,000 = 0.10 1,500>5,000 = 0.30 E₄ 0 searches e₁₀ 700>5,000 = 0.14 e₁₁ 500>5,000 = 0.10 e₁₂ 0>5,000 = 0.00 1,200>5,000 = 0.24 Total 1,900>5,000 = 0.38 2,100>5,000 = 0.42 1,000>5,000 = 0.20 5,000>5,000 = 1.00

EXAMPLE 4-9

Addition Rule for Any Two Events

Greenfield Forest Products Greenfield Forest Products manufactures lumber for large material supply centers like Home Depot and Lowe’s in the United States and Canada. A representative from Home Depot is due to arrive at the Greenfield plant for a meeting to dis- cuss lumber quality. When the Home Depot representative arrives, he will ask Greenfield’s

TABLE 4.6 Google—Addition Rule Example

Age Group

Searches per Day E₅ Under 30 E₆ 30 to 50 E₇ Over 50 Total

E₁ Ú 10 searches e₁ 200>5,000 = 0.04 e₂ 100>5,000 = 0.02 e₃ 100>5,000 = 0.02 400>5,000 = 0.08 E₂ 3 to 9 searches e₄ 600>5,000 = 0.12 e₅ 900>5,000 = 0.18 e₆ 400>5,000 = 0.08 1,900>5,000 = 0.38 E₃ 1 to 2 searches e₇ 400>5,000 = 0.08 e₈ 600>5,000 = 0.12 e₉ 500>5,000 = 0.10 1,500>5,000 = 0.30 E₄ 0 searches e₁₀ 700>5,000 = 0.14 e₁₁ 500>5,000 = 0.10 e₁₂ 0>5,000 = 0.00 1,200>5,000 = 0.24 Total 1,900>5,000 = 0.38 2,100>5,000 = 0.42 1,000>5,000 = 0.20 5,000>5,000 = 1.00

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managers to randomly select one board from the finished goods inventory for a quality check. Boards of three dimensions and three lengths are in the inventory. The following table shows the number of boards of each size and length:

Dimension

Length E2 = 2″ * 4″ E5 = 2″ * 6″ E6 = 2″ * 8″ Total

E₁ = 8 feet 1,400 1,500 1,100 4,000

E₂ = 10 feet 2,000 3,500 2,500 8,000

E₃ = 12 feet 1,600 2,000 2,400 6,000

Total 5,000 7,000 6,000 18,000

The Greenfield manager will be selecting one board at random from the inventory to show the Home Depot representative. Suppose he is interested in the probability that the board selected will be 8 feet long or a 2″ * 6″. To find this probability, he can use the following steps:

s t e p 1 Define the experiment.

One board is selected from the inventory and its dimensions are obtained.

s t e p 2 Define the events of interest.

The manager is interested in boards that are 8 feet long.

E₁ = 8 feet

He is also interested in the 2″ * 6″ dimension, so E₅ = 2″ * 6″ boards

s t e p 3 Determine the probability for each event.

There are 18,000 boards in inventory and 4,000 of these are 8 feet long, so P1E₁2 = 4,000

18,000 = 0.2222 Of the 18,000 boards, 7,000 are 2″ * 6″, so the probability is

P1E₅2 = 7,000

18,000 = 0.3889

s t e p 4 Determine whether the two events overlap, and if so, compute the joint probability.

Of the 18,000 total boards, 1,500 are 8 feet long and 2″ * 6″. Thus the joint probability is

P1E₁ and E₅2 = 1,500

18,000 = 0.0833

s t e p 5 Compute the desired probability using Probability Rule 4.

P1E₁ or E₅2 = P1E₁2 + P1E₅2 - P1E₁ and E₅2 P1E₁ or E₅2 = 0.2222 + 0.3889 - 0.0833

= 0.5278

The chance of selecting an 8-foot board or a 2″ * 6″ board is just under 0.53.

TRY EXERCISE 4-28 (pg. 184)

Addition Rule for Mutually Exclusive Events We indicated previously that when two events are mutually exclusive, both events cannot occur at the same time. Thus, for mutu- ally exclusive events,

P1E₁ and E₂2 = 0

Therefore, when you are dealing with mutually exclusive events, the Addition Rule assumes a different form, shown as Probability Rule 5.

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Figure 4.2 is a Venn diagram illustrating the application of the Addition Rule for Mutually Exclusive Events.