INDEPENDENCE in .NET

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INDEPENDENCE
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In some cases, the conditional probability of P1B A2 might equal P(B). In this special case, knowledge that the outcome of the experiment is in event A does not affect the probability that the outcome is in event B.
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EXAMPLE 2-23
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Suppose a day s production of 850 manufactured parts contains 50 parts that do not meet customer requirements. Suppose two parts are selected from the batch, but the rst part is replaced before the second part is selected. What is the probability that the second part is defective (denoted as B) given that the rst part is defective (denoted as A) The probability needed can be expressed as P1B A2. Because the rst part is replaced prior to selecting the second part, the batch still contains 850 parts, of which 50 are defective. Therefore, the probability of B does not depend on whether or not the rst part was defective. That is, P1B A2 50 850
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Also, the probability that both parts are defective is P1A B2 P1B 0 A2P1A2 a 50 50 b a b 850 850 0.0035
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2-6 INDEPENDENCE
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Table 2-4
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Parts Classi ed Surface Flaws Yes (event F) No 18 342 360 Total 20 380 400
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Defective
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Yes (event D) No Total
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EXAMPLE 2-24
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The information in Table 2-3 related surface aws to functionally defective parts. In that case, we determined that P1D F2 10 40 0.25 and P1D2 28 400 0.07. Suppose that the situation is different and follows Table 2-4. Then, P1D F2 2 40 0.05 and P1D2 20 400 0.05
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That is, the probability that the part is defective does not depend on whether it has surface aws. Also, P1F D2 2 20 0.10 and P1F2 40 400 0.10
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so the probability of a surface aw does not depend on whether the part is defective. Furthermore, the de nition of conditional probability implies that P1F D2 but in the special case of this problem P1F D2 P1D2P1F2 2 40 2 20 1 200 P1D F2P1F2
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The preceding example illustrates the following conclusions. In the special case that P1B 0 A2 P1B2, we obtain P1A B2 and P1A B2 P1A B2 P1B2 P1A2P1B2 P1B2 P1A2 P1B A2P1A2 P1B2P1A2
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These conclusions lead to an important de nition. De nition Two events are independent if any one of the following equivalent statements is true: (1) (2) (3)
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P1A B2 P1B A2 P1A B2 P1A2 P1B2 P1A2P1B2
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(2-9)
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CHAPTER 2 PROBABILITY
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It is left as a mind-expanding exercise to show that independence implies related results such as P1A B 2 P1A 2P1B 2.
The concept of independence is an important relationship between events and is used throughout this text. A mutually exclusive relationship between two events is based only on the outcomes that comprise the events. However, an independence relationship depends on the probability model used for the random experiment. Often, independence is assumed to be part of the random experiment that describes the physical system under study. EXAMPLE 2-25 A day s production of 850 manufactured parts contains 50 parts that do not meet customer requirements. Two parts are selected at random, without replacement, from the batch. Let A denote the event that the rst part is defective, and let B denote the event that the second part is defective. We suspect that these two events are not independent because knowledge that the rst part is defective suggests that it is less likely that the second part selected is defective. Indeed, P1B A2 49 849. Now, what is P(B) Finding the unconditional P(B) is somewhat dif cult because the possible values of the rst selection need to be considered: P1B2 P1B A2P1A2 P1B A 2P1A 2 149 8492150 8502 150 84921800 8502 50 850
Interestingly, P(B), the unconditional probability that the second part selected is defective, without any knowledge of the rst part, is the same as the probability that the rst part selected is defective. Yet, our goal is to assess independence. Because P1B A2 does not equal P(B), the two events are not independent, as we suspected. When considering three or more events, we can extend the de nition of independence with the following general result.
De nition The events E1, E2, p , En are independent if and only if for any subset of these events Ei1, Ei2, p , Eik , P1Ei1 Ei2 p Eik 2 P1Ei1 2 P1Ei2 2 p P1Eik 2 (2-10)
This de nition is typically used to calculate the probability that several events occur assuming that they are independent and the individual event probabilities are known. The knowledge that the events are independent usually comes from a fundamental understanding of the random experiment. EXAMPLE 2-26 Assume that the probability that a wafer contains a large particle of contamination is 0.01 and that the wafers are independent; that is, the probability that a wafer contains a large particle is