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Let A denote the event that a sample has excellent surface nish, and let B denote the event that a sample has excellent length. If a part is selected at random, determine the following probabilities: (a) P1A2 (b) P1B2 (c) P1A 2 (d) P1A B2 (e) P1A B2 (f) P1A B2 2-47. Samples of emissions from three suppliers are classi ed for conformance to air-quality speci cations. The results from 100 samples are summarized as follows: conforms 1 2 3 yes 22 25 30 no 8 5 10
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Let A denote the event that a sample is from supplier 1, and let B denote the event that a sample conforms to speci cations. If a sample is selected at random, determine the following probabilities: (a) P1A2 (b) P1B2 (c) P1A 2 (d) P1A B2 (e) P1A B2 (f) P1A B2 2-48. Use the axioms of probability to show the following: (a) For any event E, P1E 2 1 P1E2 . (b) P1 2 0 (c) If A is contained in B, then P1A2 P1B2
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Joint events are generated by applying basic set operations to individual events. Unions of events, such as A B; intersections of events, such as A B ; and complements of events, such as A , are commonly of interest. The probability of a joint event can often be determined from the probabilities of the individual events that comprise it. Basic set operations are also sometimes helpful in determining the probability of a joint event. In this section the focus is on unions of events.
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EXAMPLE 2-13
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Table 2-1 lists the history of 940 wafers in a semiconductor manufacturing process. Suppose one wafer is selected at random. Let H denote the event that the wafer contains high levels of contamination. Then, P1H2 358 940. Let C denote the event that the wafer is in the center of a sputtering tool. Then, P1C2 626 940. Also, P1H C2 is the probability that the wafer is from the center of the sputtering tool and contains high levels of contamination. Therefore, P1H C2 112 940
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The event H C is the event that a wafer is from the center of the sputtering tool or contains high levels of contamination (or both). From the table, P1H C2 872 940. An alternative calculation of P1H C2 can be obtained as follows. The 112 wafers that comprise the event H C are included once in the calculation of P(H) and again in the calculation of P(C). Therefore, P1H C2 can be found to be P1H C2 P1H2 P1C2 P1H C2 358 940 626 940 112 940
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The preceding example illustrates that the probability of A or B is interpreted as P1A B2 and that the following general addition rule applies.
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EXAMPLE 2-14
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The wafers such as those described in Example 2-13 were further classi ed as either in the center or at the edge of the sputtering tool that was used in manufacturing, and by the degree of contamination. Table 2-2 shows the proportion of wafers in each category. What is
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Table 2-1 Wafers in Semiconductor Manufacturing Classi ed by Contamination and Location Location in Sputtering Tool Contamination Low High Total Center 514 112 626 Edge 68 246 314 Total 582 358
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CHAPTER 2 PROBABILITY
Table 2-2 Wafers Classi ed by Contamination and Location Number of Contamination Particles 0 1 2 3 4 5 or more Totals
Center 0.30 0.15 0.10 0.06 0.04 0.07 0.72
Edge 0.10 0.05 0.05 0.04 0.01 0.03 0.28
Totals 0.40 0.20 0.15 0.10 0.05 0.10 1.00
the probability that a wafer was either at the edge or that it contains four or more particles Let E1 denote the event that a wafer contains four or more particles, and let E2 denote the event that a wafer is at the edge. The requested probability is P1E1 E2 2 . Now, P1E1 2 0.15 and P1E2 2 0.28. Also, from the table, P1E1 E2 2 0.04 . Therefore, using Equation 2-1, we nd that P1E1 E2 2 0.15 0.28 0.04 0.39
What is the probability that a wafer contains less than two particles or that it is both at the edge and contains more than four particles Let E1 denote the event that a wafer contains less than two particles, and let E2 denote the event that a wafer is both from the edge and contains more than four particles. The requested probability is P1E1 E2 2 . Now, P1E1 2 0.60 and P1E2 2 0.03. Also, E1 and E2 are mutually exclusive. Consequently, there are no wafers in the intersection and P1E1 E2 2 0 . Therefore, P1E1 E2 2 0.60 0.03 0.63
Recall that two events A and B are said to be mutually exclusive if A B . Then, P1A B2 0 , and the general result for the probability of A B simpli es to the third axiom of probability.
If A and B are mutually exclusive events, P1A B2 P1A2 P1B2 (2-2)
Three or More Events More complicated probabilities, such as P1A B C2 , can be determined by repeated use of Equation 2-1 and by using some basic set operations. For example, P1A B C2 P3 1A B2 C4 P1A B2 P1C2 P3 1A B2 C 4