5.6 Law of Total Probability and Bayes Rule

Suppose that you are testing a cow for mad cow disease or bovine spongiform encephalopathy (BSE). Like many medical tests, there is a chance of a false-positive. Define the event \(B\) as “cow has BSE” and event \(T\) as “cow tests positive.” Assume that we have the following probabilities: \(Pr(T|B) = 0.7\), \(Pr(T|B^C) = 0.1\), \(Pr(B) = 0.02\), and \(Pr(B^C) = 0.98\). What is \(Pr(T) = Pr(T|B) \cdot Pr(B) + Pr(T|B^C) \cdot Pr(B^C)\)? Remember from conditional probability \[Pr(T|B) = \frac{Pr(T \cap B )}{Pr(B)} \\ Pr(T|B^C) = \frac{Pr(T \cap B^C )}{Pr(B^C)}\] Using the provided probabilities, this can be expressed as \(P(T)=0.7 \cdot 0.02 + 0.1 \cdot 0.95 = 0.112\). What is the probability that a cow has BSE if it tests positive, i.e., \(P(B|T)\)? The solution to this can be written as \[Pr(B|T) = \frac{Pr(T \cap B )}{Pr(T)} = \frac{Pr(T|B) \cdot Pr(B)}{Pr(T|B) \cdot Pr(B) + Pr(T|B^C) \cdot Pr(B^C)}\] Using the provided probabilities, this can be expressed as \(Pr(B|T)=0.7 \cdot 0.02 / 0.112=0.125\). This can also be explained using a probability tree.