Practice problems for these concepts can be found at:

- Probability Solved Problems for Beginning Statistics
- Probability Supplementary Problems for Beginning Statistics

A computer disk manufacturer has three locations that produce computer disks. The Omaha plant produces 30% of the disks, of which 0.5% are defective. The Memphis plant produces 50% of the disks, of which 0.75% are defective. The Kansas City plant produces the remaining 20%, of which 0.25% are defective. If a disk is purchased at a store and found to be defective, what is the probability that it was manufactured by the Omaha plant? This type of problem can be solved using Bayes' theorem. To formalize our approach, let A_{1} be the event that the disk was manufactured by the Omaha plant, let A_{2} be the event that the disk was manufactured by the Memphis plant, and let A3 be the event that it was manufactured by the Kansas City plant. Let B be the event that the disk is defective. We are asked to find P(A_{1} | B). This probability is obtained by dividing P(A_{1} and B) by P(B).

The event that a disk is defective occurs if the disk is manufactured by the Omaha plant and is defective or if the disk is manufactured by the Memphis plant and is defective or if the disk is manufactured by the Kansas City plant and is defective. This is expressed as follows:

- B = (A

_{1}

*and*B)

*or*(A

_{2}

*and*B)

*or*(A

_{3}

*and*B) (

*4.12*)

Because the three events which are connected by or's in formula (*4.12*) are mutually exclusive, P(B) may be expressed as

- P(B) = P(A

_{1}

*and*B) + P(A

_{2}

*and*B) + P(A

_{3}

*and*B) (

*4.13*)

By using the multiplication rule, formula (*4.13*) may be expressed as

- P(B) = P(B | A

_{1}) P(A

_{1}) + P(B | A

_{2}) P(A

_{2}) + P(B | A

_{3}) P(A

_{3}) (

*4.14*)

Using formula (*4.14*), P(B) = .005 × .3 + .0075 × .5 + .0025 × .2 = .00575. That is, 0.575% of the disks manufactured by all three plants are defective. The probability P(A_{1} *and* B) equals P(B | A_{1}) P(A_{1}) = .005 × .3 = .0015. The probability we are seeking is equal to = .261. Summarizing, if a defective disk is found, the probability that it was manufactured by the Omaha plant is .261.

In using Bayes' theorem to find P(A_{1} | B) use the following steps:

Step 1: Compute P(A_{1} *and* B) by using the equation P(A_{1} *and* B) = P(B | A_{1}) P(A_{1}).

Step 2: Compute P(B) by using formula (*4.14*).

Step 3: Divide the result in step 1 by the result in step 2 to obtain P(A_{1} | B).

These same steps may be used to find P(A_{2} | B) and P(A_{3} | B).

Events like A_{1} , A_{2} , and A_{3} are called *collectively exhaustive*. They are mutually exclusive and their union equals the sample space. Bayes' theorem is applicable to any number of collectively exhaustive events.

**EXAMPLE 4.26** Using the three-step procedure given above, the probability that a defective disk was manufactured by the Memphis plant is found as follows:

Step 1: P(A_{2} and B) = P(B | A_{2}) P(A_{2}) = .0075 × .5 = .00375.

Step 2: P(B) = .005 × .3 + .0075 × .5 + .0025 × .2 = .00575.

Step 3: P(A_{2} | B) = = .652.

The probability that a defective disk was manufactured by the Kansas City plant is found as follows:

Step 1: P(A_{3} and B) = P(B | A_{3}) P(A_{3}) = .0025 × .2 = .0005.

Step 2: P(B) = .005 × .3 + .0075 × .5 + .0025 × .2 = .00575.

Step 3: P(A_{2} | B) = = .087.

Practice problems for these concepts can be found at:

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