Bayes' Theorem

A problem scenario may present information of the form $P (B | A)$ where instead we actually want $P (A | B)$ . A simple application of the Multiplication Rule for Conjunct gives us the following result:

Bayes' Theorem

Let $A_{1}, . . ., A_{k}$ be a partition of the sample space $S$ . Then for any other event $B$ for which $P (B) > 0$ :

P (A_{j} | B) = \frac{P (A_{j} \cap B)}{P (B)} = \frac{P (B | A_{j}) P (A_{j})}{\sum_{i = 1}^{k} P (B | A_{i}) P (A_{i})}

for each $j = 1, . . ., k$

Proof
The first equality comes from the Definition of Conditional Probability. The second equality rests on using the Multiplication Rule for Conjunct in the numerator, and the The Law of Total Probability in the denominator.
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Construction Example

Say the project manager for a large construction company believes the company's current bid on a project has probability $.4$ of being accepted. Say $A$ is the event the bid is accepted, so $P (A) = .4$ .
Usually supplemental information is requested, and here the project manager know that in $.7$ of cases when the company's bid was accepted, the client had asked for supplemental information. Further, in $.25$ of the cases when the bid was rejected, the client asked for supplemental information. Say $B$ be the event that information is requested. Thus $P (B | A) = .7$ and $P (B | A^{'}) = .25$ .
What is the probability of acceptance, given that supplemental information was requested? Namely, what is $P (A | B)$ ?

Proof
Apply Bayes' Theorem:

\begin{aligned} P (A | B) & = \frac{P (A \cap B)}{P (B)} \\ = \frac{P (B | A) \cdot P (A)}{P (B | A) P (A) + P (B | A^{'}) P (A^{'})} \\ = \frac{.7 \cdot .4}{.7 \cdot .4 + (1 - .4) (.25)} \\ = .65 \end{aligned}

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