www.downloadslide.com 2.9 Calculating the Probability of an Event: The Event-Composition Method 63 FIGURE 2.11 Venn diagram for events of Example 2.17 R S D F R F D F EXAMPLE 2.18 In Example 2.7 we considered an experiment wherein the birthdays of 20 randomly selected persons were recorded. Under certain conditions we found that P(A) = .5886, where A denotes the event that each person has a different birthday. Let B denote the event that at least one pair of individuals share a birthday. Find P(B). Solution The event B is the set of all sample points in S that are not in A, that is, B = A. Therefore, P(B) = 1 − P(A) = 1 − .5886 = .4114. (Most would agree that this probability is surprisingly high!) Let us refer to Example 2.4, which involves the two tennis players, and let D 1 and D 2 denote the events that player A wins the first and second games, respectively. The information given in the example implies that P(D 1 ) = P(D 2 ) = 2/3. Further, if we make the assumption that D 1 and D 2 are independent, it follows that P(D 1 ∩ D 2 ) = 2/3 × 2/3 = 4/9. In that example we identified the simple event E 1 , which we denoted AA, as meaning that player A won both games. With the present notation, E 1 = D 1 ∩ D 2 , and thus P(E 1 ) = 4/9. The probabilities assigned to the other simple events in Example 2.4 can be verified in a similar manner. The event-composition approach will not be successful unless the probabilities of the events that appear in P(A) (after the additive and multiplicative laws have been applied) are known. If one or more of these probabilities is unknown, the method fails. Often it is desirable to form compositions of mutually exclusive or independent events. Mutually exclusive events simplify the use of the additive law and the multiplicative law of probability is easier to apply to independent events.
www.downloadslide.com 64 Chapter 2 Probability A summary of the steps used in the event-composition method follows: 1. Define the experiment. 2. Visualize the nature of the sample points. Identify a few to clarify your thinking. 3. Write an equation expressing the event of interest—say, A—as a composition of two or more events, using unions, intersections, and/or complements. (Notice that this equates point sets.) Make certain that event A and the event implied by the composition represent the same set of sample points. 4. Apply the additive and multiplicative laws of probability to the compositions obtained in step 3 to find P(A). Step 3 is the most difficult because we can form many compositions that will be equivalent to event A. The trick is to form a composition in which all the probabilities appearing in step 4 are known. The event-composition approach does not require listing the sample points in S, but it does require a clear understanding of the nature of a typical sample point. The major error students tend to make in applying the event-composition approach occurs in writing the composition. That is, the point-set equation that expresses A as union and/or intersection of other events is frequently incorrect. Always test your equality to make certain that the composition implies an event that contains the same set of sample points as those in A. A comparison of the sample-point and event-composition methods for calculating the probability of an event can be obtained by applying both methods to the same problem. We will apply the event-composition approach to the problem of selecting applicants that was solved by the sample-point method in Examples 2.11 and 2.12. EXAMPLE 2.19 Solution Two applicants are randomly selected from among five who have applied for a job. Find the probability that exactly one of the two best applicants is selected, event A. Define the following two events: B: Draw the best and one of the three poorest applicants. C: Draw the second best and one of the three poorest applicants. Events B and C are mutually exclusive and A = B ∪ C. Also, let D 1 = B 1 ∩ B 2 , where B 1 = Draw the best on the first draw, B 2 = Draw one of the three poorest applicants on the second draw, and D 2 = B 3 ∩ B 4 , where B 3 = Draw one of the three poorest applicants on the first draw, B 4 = Draw the best on the second draw. Note that B = D 1 ∪ D 2 .