Sometimes we only know the conditional probabilities of an event \(A\), given that each of another collection of events \(\{B_1, ..., B_k\} \) have occurred. If we also know the probabilities of the B-events, it is possible to calculate the unconditional (i.e. marginal) probability for \(A\).
Law of total probability
If \(\{B_1, ..., B_k\} \) are a partition of the sample space,
\[ P(A) = \sum_{i=1}^{k} {P(A \mid B_i) \times P(B_i)} \](Proved in full version)
The following example illustrates this.
Question
The following table was derived from reports about contraceptive use in the USA among women aged 15 to 44 who were sexually active but not intending to get pregnant. It also shows the percentages of these women who had unintended pregnancies in their first year using each method.
Main method | % of women | % unintended pregnant |
---|---|---|
No method | 37.8% | 85% |
Pill | 17.1% | 9% |
Female sterilization | 16.5% | 0.5% |
Male sterilization | 6.2% | 0.15% |
Male condom | 10.2% | 18% |
IUD | 3.5% | 0.5% |
Withdrawal | 3.2% | 22% |
Injectable | 2.4% | 0.3% |
Vaginal ring | 1.3% | 9% |
Fertility awareness methods | 0.7% | 24% |
Patch | 0.5% | 9% |
Implant | 0.3% | 0.05% |
Other methods | 0.3% | 10% |
What is the probability that a random American woman in this age group who is sexually active but not intending to get pregnant will actually get pregnant?
(Solved in full version)