Suppose 1 in 25 adults is afflicted with a disease for which anew diagnostic tes

Question

Suppose 1 in 25 adults is afflicted with a disease for which anew diagnostic test has been developed. Given that an individualactually has the disease, a positive test result will occur withprobability .99. Given that an individual does not have thedisease, a negative test result will occur with probability.98.

Use a probability tree to answer the following questions.

A.) If a randomly selected adult is testedand the result is positive, what is the probability that theindividual has the disease?

B.) If a randomly selected adult is tested andthe result is negative, what is the probability that the individualdoes not have the disease?

Explanation / Answer

This is a Bayes' Theorem question (though you may not haveheard it by that name yet). . By definition, Pr(A&B) = Pr(A|B)*Pr(B) . Likewise, Pr(A&B) = Pr(B|A)*Pr(A) . Pr(A|B)Pr(B) = Pr(A&B) = Pr(B|A)Pr(A) . Pr(A|B) = Pr(B/A) * Pr(A) / Pr(B) .       where Pr(A|B) is theprobability of A, conditioned on B's having happened. . You have some prior belief about an event A's happening (suchas having the disease). In thiscase, Pr(A)= (1/25). If we didn't know anythingabout test results, this is the probability we'd assign. Now,however, we get a SIGNAL, the additional information that event B(a positive test result) has occurred. How should we changeour belief in the probability of A, to condition on the fact that Bhas occurred? . Pr(A|B) = Pr(B|A) * Pr(A) / Pr(B) . We have to find THREE pieces of information. Pr(B|A): what's the probability that an individual with thedisease will test positive. The problem tells you this. Pr(A): what's the underlying probability that a randomlychosen individual has the disease. The problem tells youthis. Pr(B): what's the underlying probability that a randomlychosen individual gets a positive test. The problem does NOTtell you this directly, but gives you enough information to work itout. . Suppose we had 100,000 individuals, picked at random. How many of them should have the disease? Of those who havethe disease, how many should test positive? Of those whodo NOT have the disease, how many should test positive? Divide the total number of positive tests by 100,000 and you havePr(B). (You could set this up as a weighted average instead,if you're comfy with weighted averages.) . Once you've done the 1st part of the question (probabilityhaving the disease|positive test), the 2nd part (probability of NOThaving the disease|negative test) works just the same way. Note that the probability of a negative test will be one minus theprobability of a positive test, so you've already done most of thework for this part. . EXAMPLE: suppose 5% of all tax returns contain a majorerror. The IRS has a screening mechanism that pops up a redflag for scrutiny. 90% of erroneous returns pop the flag; 4%of non-erroneous returns pop the flag. What's the probabilitythat a flagged return has a major error? Let A = return has an error and B = return has beenflagged Pr(A) = 0.05 Pr(B|A) = 0.9 Pr(B):    If we had 100,000 returns, 95,000 would benon-erroneous, of which 4%, or 3800, will pop a flag.    5,000 returns will be erroneous, of which 90%, or4500, will pop a flag.    Pr(B) = (3800+4500)/100,000 = 0.083 [Note:check to make sure that Pr(B) lies in between Pr(B|A) and Pr(B|notA). It does in this case, so we're good.] . Pr(A|B) = 0.9 * 0.05 / 0.083 = 54.2% ------------------------------------------ I didn't notice the part about using a probability tree, butthe math works out the same. . Draw a tree that starts with the node for (Have Disease, Don'tHave Disease). The Have Disease branch has p=(1/25) andthe Don't Have Disease branch has p=(24/25). At the end ofeach branch, add a node for Tests Positive/Tests Negative, and fillin the probabilities. . At the end of the tree, you have the nodes for     Have Disease, Test Positive =Pr(A&B)     Have Disease, Test Negative = Pr(A &NOT B)     Don't Have Disease, Test Positive = Pr(NOTA & B)     Don't Have Disease, Test Negative = Pr(NOTA & NOT B) and you'll have the probabilities of each outcome. Then    Pr(Have Disease|Test Positive) = Pr(A&B) /[Pr(A&B) + Pr(NOT A & B)] . which gives the same answer as the Bayes' Theorem approach(and requires the exact same calculations, just in a slightlydifferent order).

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