Question

Question:State Bayes' theorem and use it to find

if \(p(E\mid F) = 1/3,p(E\mid \bar F) = 1/4\), and \(p(F) = 2/3\), where \(E\) and \(F\) are events from a sample space \(S\).

Short answer

Answer

The resultant answer is Bayes’ theorem: \(P(F\mid E) = \frac{{P(E\mid F)P(F)}}{{P(E\mid F)P(F) + P(E\mid \bar F)P(\bar F)}}\) \(\begin{aligned}{}P(F\mid E)= \frac{8}{{11}}\\ \approx 0.7273\end{aligned}\).

Step-by-step solution

Given data 

\(E\) and \(F\) are the given events.

Concept of Bayes’ theorem and complement rule

Bayes' theorem

Complement rule:\(P(\bar E) = 1 - P(E)\).

Simplify the expression

Use the Bayes’ theorem and simplify the expression

\(P(E\mid F) = \frac{1}{3}\)

\(P(E\mid \bar F) = \frac{1}{4}\)

\(P(F) = \frac{2}{3}\)

Use the complement rule:

\(P(\bar F) = 1 - P(F) = 1 - \frac{2}{3} = \frac{1}{3}\)

Use Bayes' theorem:

\(\begin{aligned}{}P(F\mid E) &= \frac{{P(E\mid F)P(F)}}{{P(E\mid F)P(F) + P(E\mid \bar F)P(\bar F)}}\\P(F\mid E) &= \frac{{\frac{1}{3} \cdot \frac{2}{3}}}{{\frac{1}{3} \cdot \frac{2}{3} + \frac{1}{4} \cdot \frac{1}{3}}}\\P(F\mid E) &= \frac{{\frac{2}{9}}}{{\frac{{11}}{{36}}}}\end{aligned}\)

\(\begin{aligned}{}P(F\mid E) = \frac{8}{{11}}\\P(F\mid E) \approx 0.7273\end{aligned}\)