Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 9F / a

Table of contents
- Solution a

Solution

By the definition of conditional probability,

P (B_{i} ∣ A) = \frac{P ( A \cap B _{i} )}{P ( A )} = \frac{P ( A ∣ B _{i} ) P ( B _{i} )}{P ( A )} .

Since the

B_{j}

form a disjoint partition of the sample space, the law of total probability gives

P (A) = \sum_{j} P (A ∣ B_{j}) P (B_{j})

. Substitution proves Bayes' theorem in the required form:

P (B_{i} ∣ A) = \frac{P ( A ∣ B _{i} ) P ( B _{i} )}{\sum _{j} P ( A ∣ B _{j} ) P ( B _{j} )} .

The hypotheses make the denominator positive and each conditioning probability well defined.

 Ancestors (11)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version