Multiplication Rule

4.2 The Rules of Probability

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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)

178 Chapter 4

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Introduction to Probability

BUSINESS APPLICATION

Multiplication Rule

Hong Kong Fireworks The Hong Kong Fireworks Company is a manufacturer of fire- works used by cities, fairs, and other commercial establishments for large-scale fireworks displays. The company uses two suppliers of material in making a particular product. The materials from the two suppliers are intermingled in the manufacturing process. When a case of fireworks is being made, the material is pulled randomly from inventory without regard to which company made it. Recently, a customer ordered two products. At the time of assembly, the material inventory contained 30 units of MATX and 50 units of Quinex. What is the prob- ability that both fireworks products ordered by this customer will have MATX material?

To answer this question, we must recognize that two events are required to form the desired outcome:

E₁ = Event: MATX material in first product E₂ = Event: MATX material in second product

The probability that both fireworks products contain MATX material is written as P1E₁ and E₂2. The key word here is and, as contrasted with the Addition Rule, in which the key word is or. The and signifies that we are interested in the joint probability of two events, as noted by P1E₁ and E₂2. To find this probability, we employ Probability Rule 8:

P1E₁ and E₂2 = P1E₁2P1E₂E₁2

We start by assuming that each unit of material in the inventory has the same chance of being selected for assembly. For the first fireworks product,

P(E₁) = Number of MATX units

Number of firework materials units in inventory

= 30

80 = 0.375

Then, because we are not replacing the first firework material, we find P1E₂E₁2 by P(E₂E₁) = Number of remaining MATX units

Number of remaining firework materials units

= 29

79 = 0.3671 Now, by Probability Rule 8,

P1E₁ and E₂2 = P1E₁2P1E₂E₁2 = 10.375210.36712

= 0.1377

Therefore, there is a 13.77% chance the two fireworks products will contain the MATX material.

Using a Tree Diagram

BUSINESS APPLICATION

Multiplication Rule

Hong Kong Fireworks (continued) A tree diagram can be used to display the situation facing Hong Kong Fireworks. The company uses material from two suppliers, which is inter- mingled in the inventory. Recently a customer ordered two products and found that both con- tained the MATX material. Assuming that the inventory contains 30 MATX and 50 Quinex units, to determine the probability of both products containing the MATX material you can use a tree diagram. The two branches on the left side of the tree in Figure 4.4 show the possi- ble material options for the first product. The two branches coming from each of the first branches show the possible material options for the second product. The probabilities at the

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far right are the joint probabilities for the material options for the two products. As we deter- mined previously, the probability that both products will contain a MATX unit is 0.1377, as shown on the top right on the tree diagram.

We can use the Multiplication Rule and the Addition Rule in one application when we determine the probability that two products will have different materials. Looking at Figure 4.4, we see there are two ways this can happen:

P31 MATX and Quinex 2 or 1 Quinex and MATX 24 = ?

If the first product is a MATX and the second one is a Quinex, then the first cannot be a Quinex and the second a MATX. These two events are mutually exclusive and, there- fore, Probability Rule 5 can be used to calculate the required probability. The joint prob- abilities (generated from the Multiplication Rule) are shown on the right side of the tree.

To find the desired probability, using Probability Rule 5, we can add the two joint probabilities:

P31 MATX and Quinex2 or 1Quinex and MATX 24 = 0.2373 + 0.2373 = 0.4746

The chance that a customer buying two products will get two different materials is 47.46%.

P(MATX and MATX) = 0.375 0.3671 = 0.1377 MATX

P = 29/79 = 0.3671 Quinex

P = 50/79 = 0.6329 MATX

P = 30/80 = 0.375

Quinex

P = 50/80 = 0.625

Product 1 Product 2

P(MATX and Quinex) = 0.375 0.6329 = 0.2373 P(Quinex and MATX) = 0.625 0.3797 = 0.2373

P(Quinex and Quinex) = 0.625 0.6203 = 0.3877 MATX

P = 30/79 = 0.3797 Quinex

P = 49/79 = 0.6203 FIGURE 4.4 Tree Diagram

for the Fireworks Product Example

Multiplication Rule for Independent Events When we determined the probability that two products would have MATX material, we used the general multiplication rule (Rule 8). The general multiplication rule requires that conditional probability be used because the probability associated with the second product depends on the material selected for the first product. The chance of obtaining a MATX was lowered from 30/80 to 29/79, given that the first material was a MATX.

However, if the two events of interest are independent, the imposed condition does not alter the probability, and the Multiplication Rule takes the form shown in Probability Rule 9.

The joint probability of two independent events is simply the product of the probabilities of the two events. Rule 9 is the primary way that you can determine whether any two events are independent. If the product of the probabilities of the two events equals the joint probability, then the events are independent.

Probability Rule 9: Multiplication Rule for Independent Events For independent events E₁ and E₂,

P1E₁ and E₂2 = P1E₁2P1E₂2 (4.12)

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Introduction to Probability

EXAMPLE 4-12

Using the Multiplication Rule and the Addition Rule

Christiansen Accounting Christiansen Accounting prepares tax returns for individu- als and companies. Over the years, the firm has tracked its clients and has discovered that 12% of the individual returns have been selected for audit by the Internal Revenue Service. On one particular day, the firm signed two new individual tax clients. The firm is interested in the probability that at least one of these clients will be audited. This proba- bility can be found using the following steps:

s t e p 1 Define the experiment.

The IRS randomly selects a tax return to audit.

s t e p 2 Define the possible outcomes.

For a single client, the following outcomes are defined:

A = Audit N = No audit For each of the clients, we define the outcomes as

Client 1: A₁; N₁ Client 2: A₂; N₂

s t e p 3 Define the overall event of interest.

The event that Christiansen Accounting is interested in is E = At least one client is audited

s t e p 4 List the outcomes for the events of interest.

The possible outcomes for which at least one client will be audited are as follows:

E₁: A₁ A₂ both are audited

E₂: A₁ N₂

only one client is audited E₃: N₁ A₂

s t e p 5 Compute the probabilities for the events of interest.

Assuming the chances of the clients being audited are independent of each other, probabilities for the events are determined using Probability Rule 9 for independent events:

P1E₁2 = P1A₁ and A₂2 = 0.12 * 0.12 = 0.0144 P1E₂2 = P1A₁ and N₂2 = 0.12 * 0.88 = 0.1056 P1E₃2 = P1N₁ and A₂2 = 0.88 * 0.12 = 0.1056

s t e p 6 Determine the probability for the overall event of interest.

Because events E₁, E₂, and E₃ are mutually exclusive, compute the probability of at least one client being audited using Rule 5, the Addition Rule for Mutually Exclusive Events:

P1E₁ or E₂ or E₃2 = P1E₁2 + P1E₂2 + P1E₃2

= 0.0144 + 0.1056 + 0.1056

= 0.2256

The chance of one or both of the clients being audited is 0.2256.

TRY EXERCISE 4-30 (pg. 184)