176 CHAPTER 4 Probability INTERPRETATION For the data given in this example, a randomly selected subject has a 1% chance of cancer, but for a randomly selected subject given a test with a positive result, the chance of cancer increases to 7.48%. Based on the data given in this example, a positive test result should not be devastating news, because there is still a good chance that the test is wrong. ■ Among the 990 subjects without cancer, 10% get positive test results, so 10% of the 990 cancer-free subjects in the second row get positive test results. See the entry of 99 in the second row. ■ For the 990 subjects in the second row, 99 test positive, so the other 891 must test negative. See the entry of 891 in the second row. ■ Among the 10 subjects with cancer in the first row, 80% of the test results are positive, so 80% of the 10 subjects in the first row test positive. See the entry of 8 in the first row. ■ The other 2 subjects in the first row test negative. See the entry of 2 in the first row. To find P(C positive test result), see that the first column of values includes the positive test results. In that first column, the probability of randomly selecting a subject with cancer is 8>107 or 0.0748, so P(C positive test result) = 0.0748. YOUR TURN. Do Exercise 19 “Positive Predictive Value.” TABLE 4-2 Test Results Positive Test Result (Test shows cancer.) Negative Test Result (Test shows no cancer.) Total Cancer 8 (True Positive) 2 (False Negative) 10 No Cancer 99 (False Positive) 891 (True Negative) 990 The solution in Example 4 is not very difficult. Another approach is to compute the probability using this formula commonly given with Bayes’ theorem: P1A B2 = P1A2 # P1B A2 3P1A2 # P1B A24 + 3P1A2 # P1B A24 If we replace A with C and replace B with “positive,” we get this solution for Example 4: P1C positive2 = P1C2 # P1positive C2 P1C2 # P1positive C2 + P1C2 # P1positive C2 = 0.01# 0.80 10.01# 0.802 + 10.99# 0.102 = 0.0748 Study Results Here is a truly fascinating fact: When 100 physicians were given the information in Example 4, 95 of them estimated P(C positive) to be around 0.70 to 0.80, so they were wrong by a factor of 10. Physicians are extremely intelligent, but here they likely suffered from confusion of the inverse. The given rate of 80% for positive test results among those who are true positives implies that P(positive C) = 0.80, but this is very different from P(C positive). The physicians would have done much better if they had seen the given information in the form of a table like Table 4-2. Coincidences More Likely Than They Seem Evelyn Evans won $3.9 million in the New Jersey lottery, then she won another $1.5 million only 4 months later. The New York Times reported that the chance of that happening was only 1 in 17 trillion. But that likelihood is misleading because it represents the chance of Evelyn Evans winning with only one ticket purchased in each of the two specific lottery drawings. A better question would be this: What is the chance of someone somewhere winning a lottery twice? Statisticians George McCabe and Steve Samuels found that over a 7-year span, there is a 53% chance of at least one past lottery winner getting lucky with another win. The chance of “1 in 17 trillion” is sensational, but the more realistic chance is 53%. E w in J t a m
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