Bayes' theorem and total probability problem. Suppose 5 men out of 100 and 25 women out of 1000 are good orator. An orator is chosen at random. **_Find the probability that a male person is selected_**. Assume that there are equal number of men and women. **My Approach** : Finding the probability of choosing a male given that a good orator is chosen using Bayes' theorem which turns out to be $\frac{2}{3}$ Finding probability of choosing male given that not a good orator is chosen using Bayes' theorem which turns out to be $\frac{38}{77}$ **Problem** : Using total probability if I add them, the answers becomes larger than 1 which should not be the case ($\frac{2}{3}+\frac{38}{77}=1.160176$). Should I instead multiply them, counting the required probability as an "AND" case of both? (I think the events are independent hence multiplication)

Consider the events:

* $X$ = a male is chosen
* $X^c$= a female is chosen
* $O$ = a good orator is chosen
* $O^c$ = a bad orator is chosen



Your assumptions are $p(X)=p(X^c)=\frac{1}{2}$, $p(O|X)=0.05$ and $p(O|X^c)=0.025$. The probability of selecting a male conditional on choosing a good orator is

\begin{equation*} p(X|O) = \dfrac{p(O|X)p(X)}{p(O|X)p(X)+p(O|X^c)p(X^c)} \end{equation*}

The probability of selecting a male conditional on not choosing a good orator is \begin{equation*} p(X|O^c) = \dfrac{p(O^c|X)p(X)}{p(O^c|X)p(X)+p(O^c|X^c)p(X^c)} \end{equation*}

As you see, there is no reason for the condition $p(X|O)+p(X|O^c)=1$ to hold. The valid condition is \begin{equation*} p(X|O)p(O) + p(X|O^c)p(O^c) = p(X) \end{equation*} In other words, the total probability of selecting a male equals the sum of conditional probabilities multiplied by the probability of the conditioning events.