Conditional Probability in Java

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The portion of this area corresponding to the expert is 1 65
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P(E|A) =
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1 64
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The shaded areas in Figure 33 are in reality equal, but they do not appear to be equal
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Recall that the statement P(A) = P(A and E) + P(A and E) is often known as the Law of Total Probability We add the probabilities here of mutually exclusive events, namely, A and E and A and E (represented by the nonoverlapping rectangles) The law can be extended to more than two mutually exclusive events
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EXAMPLE 34
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The Cancer Test
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We now return to the medical example at the beginning of this chapter We interpret the statement that the cancer test is accurate for 95% of patients with cancer and 95% accurate for patients without cancer as that P(T + |C) = 095 and P(T + |C) = 005, where C indicates the presence of cancer and T + means that the test indicates a positive result or the presence of cancer P(T + |C) is known as the false positive rate for the test since it produces a positive result for noncancer patients Supposing that a small percentage of the population has cancer, we assume in this case that P(C) = 0005 This assumption will prove crucial in our conclusions A picture will clarify the situation, although, again, the small probabilities involved force us to exaggerate the picture somewhat Figure 34 shows along the horizontal axis the probability that a person has cancer Along the vertical axis are the probabilities that a person shows a positive test for each of the two groups of patients
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0 0005
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Some Notation It is clear that the probability that a person shows a positive test is P(T + ) = 095 0005 + 005 0995 = 00545 The portion of this area corresponding to the people who actually have cancer is then P(C|T + ) = 095 0005 095 0005 + 005 0995 000475 = 0087 00545
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This is surprisingly low We emphasize, however, that the test should not be relied upon alone; one should have other indications of the disease as well We also note here that the probability that a person testing positive actually has cancer highly depends upon the true proportion of people in the population who are actually cancer patients Let us suppose that this true proportion is r, so that r represents the incidence rate of the disease Replacing the proportion 0005 by r and the proportion 0995 by 1 r in the above calculation, we nd that the proportion of people who test positive and actually have the disease is P(C|T + ) = This can be simpli ed to P(C|T + ) = 19r 095 r = 005 + 090 r 1 + 18r r 095 r 095 + (1 r) 005
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A graph of this function is shown in Figure 35 We see that the test is quite reliable when the incidence rate for the disease is large Most diseases, however, have small incidence rates, so the false positive rate for these tests is a very important number Now suppose that the test has probability p indicating the disease among patients who actually have the disease and that the test indicates the presence of the disease with probability 1 p among patients who do not have the disease p = 095 in our example It is also interesting
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1 08 P[C|T +] 06 04 02 0 0 02 04 r 06 08 1
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1 P(C|T+) 075 05 025 0 0 025 05 r 075 1 0 025 05 075 1
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to examine the probability P(C|T + ) as a function of both the incidence rate of the disease, r, and p Now P(C|T + ) = r p r p + (1 r) (1 p)
The surface showing this probability as a function of both r and p is shown in Figure 36 Clearly, the accuracy of the test increases as r and p increase Ours has a low probability since we are dealing with the lower left corner of the surface