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)= _i=13P(A_i)P(B A_i) and P(A_j B)= P(A_j)P(B A_j)P(B), provided P(B)>0. **Source** | **Share of output** | **Defect probability** | **Share of all output that is defective** Factory A | 0.6 | 0.02 | 0.012 Factory B | 0.4 | 0.05 | 0.020 Given that an item is defective, P(Factory B)= 0.0200.012+0.020=0.625. The denominator includes every route to the observed defect.

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.

Probability

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Bayes’ Theorem

Bayes’ Theorem