3.3: Bayes' Rule
- Page ID
- 47234
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Consider a composite event \(M\) and a simple event \(A_i\). We know from conditional probability from the previous section that
\begin{align*} p(A_i|M) = \dfrac{p(A_i \cap M)}{p(M)} \\[4pt] p(M|A_i) = \dfrac{p(A_i \cap M)}{p(A_i)}, \end{align*}
and if we eliminate the denominator on the right-hand side, we find that
\begin{align*} p(M|A_i) = \dfrac{p(A_i|M) \, p(M)}{p(A_i)} \\[4pt] p(A_i|M) = \dfrac{p(M|A_i) \, p(A_i)}{p(M)}. \end{align*}
The second of these is most interesting - it gives the probability of a simple event, conditioned on the composite event, in terms of the composite event conditioned on the simple one! Recalling our above formula for \(p(M)\), we thus derive Bayes’ rule:
\[ p(A_i|M) = \dfrac{p(M|A_i) \, p(A_i)}{p(M|A_1) \, p(A_1) \, + \, . \, . \, . + \, p(M|A_n) \, p(A_n)}. \]
Here is an example of its use.
Consider a medical test that is 99% accurate - it gives a negative result for people who do not have the disease 99% of the time, and it gives a positive result for people who do have the disease 99% of the time. Only one percent of the population has this disease. Joe just got a positive test result: What is the probability that he has the disease?
Solution
The composite event \(M\) is that he has the disease, and the simple events are that he tested positive \((+)\) or he tested negative \((-)\). We apply
\begin{align*} p(M|+) &= \dfrac{p(+|M) \, p(M)}{p(+)} \\[4pt] &= \dfrac{p(+|M) \, p(M)} {p(+|M) \, p(M) + p(+|\bar{M}) \, p(\bar{M})} \\[4pt] &= \dfrac{0.99 \times 0.01} {0.99 \times 0.01 + 0.01 \times 0.99} \\[4pt] &= 1/2. \end{align*}
This example is not well appreciated by many healthcare consumers!
Here is another example, without so many symmetries.
Box A has nine red pillows in it and one white. Box B has six red pillows in it and nine white. Selecting a box at random and pulling out a pillow at random gives the result of a red pillow. What is the probability that it came from Box A?
Solution
\(M\) is the composite event that it came from Box A; the simple event is that a red pillow was collected \((R)\). We have
\begin{align*} p(M|R) &= \dfrac{p(R|M) \, p(M)}{p(R)} \\[4pt] &= \dfrac{p(R|M) \, p(M)} {p(R|M) \, p(M) + p(R|\bar{M}) \, p(\bar{M})} \\[4pt] &= \dfrac{0.9 \times 0.5} {0.9 \times 0.5 + 0.4 \times 0.5} \\[4pt] &= 0.692. \end{align*}