Problem
According to Bayes' Theorem, the probability of event A, given that event B has occurred, is as follows. \[ P(A \mid B)=\frac{P(A) \cdot P(B \mid A)}{P(A) \cdot P(B \mid A)+P\left(A^{\prime}\right) \cdot P\left(B \mid A^{\prime}\right)} \] Use Bayes' Theorem to find $P(A \mid B)$ using the probabilities shown below. \[ P(A)=\frac{2}{3}, P\left(A^{\prime}\right)=\frac{1}{3}, P(B \mid A)=\frac{1}{7}, \text { and } P\left(B \mid A^{\prime}\right)=\frac{7}{10} \] The probability of event $A$, given that event $B$ has occurred, is $P(A \mid B)=$ (Round to the nearest thousandth as needed.)
Solution
Step 1 ：Given the probabilities $P(A)=\frac{2}{3}$, $P(A')=\frac{1}{3}$, $P(B|A)=\frac{1}{7}$, and $P(B|A')=\frac{7}{10}$.
Step 2 ：We can use Bayes' Theorem to find $P(A|B)$, which is defined as $P(A|B)=\frac{P(A) \cdot P(B|A)}{P(A) \cdot P(B|A)+P(A') \cdot P(B|A')}$.
Step 3 ：Substitute the given probabilities into the formula, we get $P(A|B)=\frac{\frac{2}{3} \cdot \frac{1}{7}}{\frac{2}{3} \cdot \frac{1}{7}+\frac{1}{3} \cdot \frac{7}{10}}$.
Step 4 ：Simplify the above expression to get $P(A|B)=0.29$.
Step 5 ：Final Answer: The probability of event $A$, given that event $B$ has occurred, is $P(A|B)=\boxed{0.29}$.
