Bayes’ Theorem

HL only. Bayes’ theorem reverses a conditional probability. Use no more than three mutually exclusive, exhaustive source events in the prescribed course.

For source events \(A_1,A_2,A_3\) and observed event \(B\), \(P(B)=\sum\limits_{i=1}^{3}P(A_i)P(B\mid A_i)\) and \(P(A_j\mid B)=\dfrac{P(A_j)P(B\mid A_j)}{P(B)}\), provided \(P(B)>0\).

Given that an item is defective, \(P(\text{Factory B}\mid\text{defect})=\dfrac{0.020}{0.012+0.020}=0.625\). The denominator includes every route to the observed defect.

For each possible source event, multiply its prior probability by the conditional probability of the observed evidence. Add these joint probabilities to obtain the total probability of the evidence. Divide the relevant joint probability by that same total to find the updated source probability.

The source events must be mutually exclusive and exhaustive; here use at most three. The evidence must have positive probability. Check that the posterior probabilities over all source events sum to one. A large probability of evidence given a source does not by itself imply a large probability of that source given the evidence.