Conditional Probability

4.2 The Rules of Probability

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

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Introduction to Probability

gender. For example, suppose the company knows a user is female and wants to know the chances this woman will spend between 20 and 40 hours a month on the Internet. Let

E₂ = 5e₃, e₄6 = Event: Person uses services 20 to 40 hours per month E₄ = 5e₁, e₃, e₅6 = Event: User is female

A marketing analyst needs to know the probability of E₂ given E₄. One way to find the desired probability is as follows:

1. We know E₄ has occurred (customer is female). There are 850 females in the survey.

2. Of the 850 females, 300 use Internet services 20 to 40 hours per month.

3. Then,

P1E₂E₄2 = 300 850 = 0.35 However, we can also apply Probability Rule 6, as follows:

P1E₂E₄2 = P1E₂ and E₄2 P1E₄ 2

Table 4.8 shows the relative frequencies of interest. From Table 4.8, we get the joint probability

P1E₂ and E₄2 = 0.12 and

P1E₄2 = 0.34 Then, applying Probability Rule 6, we have

P1E₂E₄2 = 0.12 0.34 = 0.35

TABLE 4.7 Joint Frequency Distribution for R.C. Minch Networking Gender

Hours per Month E₄ Female E₅ Male Total

E₁ 6 20 e₁ 450 e₂ 500 950

E₂ 20 to 40 e₃ 300 e₄ 800 1,100

E₃ 7 40 e₅ 100 e₆ 350 450

Total 850 1,650 2,500

TABLE 4.8 Joint Relative Frequency Distribution for R.C. Minch Networking Gender

Hours per Month E₄ Female E₅ Male Total

E₁ 6 20 e₁ 450>2,500 = 0.18 e₂ 500>2,500 = 0.20 950>2,500 = 0.38 E₂ 20 to 40 e₃ 300>2,500 = 0.12 e₄ 800>2,500 = 0.32 1,100>2,500 = 0.44 E₃ 7 40 e₅ 100>2,500 = 0.04 e₆ 350>2,500 = 0.14 450>2,500 = 0.18 Total 850>2,500 = 0.34 1,650>2,500 = 0.66 2,500>2,500 = 1.00

EXAMPLE 4-10

Computing Conditional Probabilities

Retirement Planning Most financial publications suggest that the older the investor is, the more conservative his or her investment strategy should be. For example, younger inves- tors might hold more in stocks, while older investors might hold more in bonds. Suppose the following table was created from a survey of investors. The table shows the number of people in the study by age group and percent of retirement funds in the stock market.

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Percent of Retirement Investments in the Stock Market Age of

Investor E₅

* 5% E₆

5% * 10% E₇

10% * 30% E₈

30% * 50% E₉

50% or more Total

E₁ 630 70 240 270 80 55 715

E₂ 30 6 50 90 300 630 1,120 1,420 3,560

E₃ 50 6 65 110 305 780 530 480 2,205

E₄ 65+ 200 170 370 260 65 1,065

Total 470 1,015 2,050 1,990 2,020 7,545

We are interested in knowing the probability that someone 65 or older will have 50%

or more of retirement funds invested in the stock market. Assuming the data collected in this study reflect the population of investors, we can find this conditional probability using the following steps:

s t e p 1 Define the experiment.

A randomly selected person age 65 or older has his or her portfolio analyzed for percent of retirement funds in the stock market.

s t e p 2 Define the events of interest.

In this case, we are interested in two events:

E₄ = At least 65 years old E₉ = 50, or more in stocks

s t e p 3 Define the probability statement of interest.

We are interested in

P1E₉E₄2 = Probability of 50, or more stocks given at least 65 years

s t e p 4 Convert the data to probabilities using the relative frequency assessment method.

We begin with the event that is given to have occurred 1E₄2. A total of 1,065 people in the study were at least 65 years of age. Of the 1,065 people, 65 had 50% or more of their retirement funds in the stock market.

P1E₉E₄2 = 65

1,065 = 0.061

Thus, the conditional probability that someone at least 65 will have 50% or more of retirement assets in the stock market is 0.061. This value can also be found using Probability Rule 6 as shown in Step 5.

s t e p 5 Use Probability Rule 6 to find the conditional probability.

You can also apply Probability Rule 6 to find the probability of a person having 50% or more of retirement assets in the stock market given that he or she is at least 65 years old.

P(E₉E₄) = P(E₉ and E₄) P(E₄)

The frequency assessment method is used to find the probability that the indi- vidual is 65 years or older:

P(E₄) = 1,065

7,545 = 0.1412

The joint probability of having 50% or more in stocks and being 65 years or older is P(E₉ and E₄) = 65

7,545 = 0.0086 Then, using Probability Rule 6, we get

P(E₉E₄) = P(E₉ and E₄)

P(E₄) = 0.0086

0.1412 = 0.061

TRY EXERCISE 4-34 (pg. 185)

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Introduction to Probability

Tree Diagrams Another way of organizing the events of an experiment that aids in the calculation of probabilities is the tree diagram.

BUSINESS APPLICATION

Using Tree Diagrams

R.C. Minch Networking (continued) Figure 4.3 illustrates the tree diagram for R.C.

Minch Networking, the Internet service provider discussed earlier. Note that the branches at each node in the tree diagram represent mutually exclusive events. Moving from left to right, the first two branches indicate the two customer types (male and female—mutually exclusive events). Three branches grow from each of these original branches, representing the three possible categories for Internet use. The probabilities for the events male and female are shown on the first two branches. The probabilities shown on the right of the tree are the joint probabilities for each combination of gender and hours of use. These figures are found using Table 4.8, which was shown earlier. The probabilities on the branches following the male and female branches showing hours of use are conditional probabilities. For example, we can find the probability that a male customer 1E₅2 will spend more than 40 hours on the Internet 1E₃2 by

P(E₃E₅) = P(E₃ and E₅)

P(E₅) = 0.14

0.66 = 0.2121

Conditional Probability for Independent Events We previously discussed that two events are independent if the occurrence of one event has no bearing on the probability that the second event occurs. Therefore, when two events are independent, the rule for condi- tional probability takes a different form, as indicated in Probability Rule 7.

Probability Rule 7: Conditional Probability for Independent Events For independent events E₁ and E₂,

P1E₁E₂2 = P1E₁2 P1E₂2 7 0 and

P1E₂E₁2 = P1E₂2 P1E₁2 7 0

(4.10)

As Probablity Rule 7 shows, the conditional probability of one event occurring, given that a second independent event has already occurred, is simply the probability of the event occurring.

P(E1 and E5) = 0.20 P(E₁|E₅)

< 20 hours

P(E₃|E₅)

> 40 hours

= 0.4848 P(E₂|E₅) 20 to 40 hours Male

P(E5) = 0.66

Female P(E₄) = 0.34

P(E₂ and E₅) = 0.32

P(E₃ and E₅) = 0.14

P(E₁ and E₄) = 0.18

P(E₂ and E₄) = 0.12

P(E₃ and E₄) = 0.04

= 0.3529 P(E2|E4) 20 to 40 hours P(E₁|E₄)

< 20 hours

= 0.1176 P(E3|E4)

> 40 hours

= 0.3030

= 0.2121

= 0.5294 FIGURE 4.3 Tree Diagram

for R.C. Minch Networking

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EXAMPLE 4-11

Checking for Independence

Greenfield Forest Products (continued) In Example 4-9, the manager at the Green- field Forest Products Company reported the following data on the boards in inventory:

Dimension

E₄ E₅ E₆

Length 2″ * 4″ 2″ * 6″ 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

He will be selecting one board at random from the inventory to show a visiting customer.

Of interest is whether the length of the board is independent of the dimension. This can be determined using the following steps:

s t e p 1 Define the experiment.

A board is randomly selected and its dimensions determined.

s t e p 2 Define one event for length and one event for dimension.

Let E₂ = Event that the board is 10 feet long and E₅ = Event that the board is 2″ * 6″.

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

P(E₂) = 8,000

18,000 = 0.4444 and P(E₅) = 7,000

18,000 = 0.3889

s t e p 4 Assess the joint probability of the two events occurring.

P(E₂ and E₅) = 3,500

18,000 = 0.1944

s t e p 5 Compute the conditional probability of one event given the other using Probability Rule 6.

P(E₂E₅) = P(E₂ and E₅)

P(E₅) = 0.1944

0.3889 = 0.50

s t e p 6 Check for independence using Probability Rule 7.

Because P1E₂E₅2= 0.50 7 P1E₂2= 0.4444, the two events, board length and board dimension, are not independent.

TRY EXERCISE 4-42 (pg. 186)