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)
 A: 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.
A: There is no reason to add or multiply them.  What would that represent? 
You could take a weighted combination such as  $\dfrac{2}{3}\times \dfrac{75}{2000} + \dfrac{38}{77} \times \dfrac{2000-75}{2000} = \dfrac12$ to check your answers are consistent with your assumption of equal numbers of men and women, but that is about it.
A: as its a average, you should consider the concept of conglomerability except in very special cases (some paradoxical cases) Which often involve issues surrounding some situations where. Then the absolute probability of $A$ is generally such that$(A)$:
$$\text{'Conglomerability'}\quad\text{Min}(P(A|\neg C),P(A|C)\,\,) \leq P(A) \leq \text{Min}(P(A|\neg C),P(A|C))$$
$$\text{As}\, P(A)\, \text{is a  convex combination of 2 numbers}: \,P(A|C), P(A|\neg C),\,\text{where,both}, (\,P(A|\neg C),P(A|C)\,)\in [0,1]$$
$$\text{Let P(C)=x},\text{thus,as,}$$,
$$(1)'['(\neg C \land C) \land  \equiv \bot' \rightarrow C \bot \neg  C,\,\text{that,is, C,and $\neg C$ are mutually,exclusive}$$
$$(2) ( (C\vee \neg C)=\top, \text{C and C are mutually exhaustive}$$ 
$$(3) \text{as P(C) is defined, so 1-P(C), presumably}$$
$$ (4)\text{then, as}\quad P(\neg C)+P(C))=1 \leftrightarrow [P(\neg C)=[1-P(  C)]=[1-x\,]\,]\rightarrow P(\neg C)=1-x$$
$$(5)\leftrightarrow (P(C),P(\neg C))=(x,1-x),\text{where,} (P(C),P(\neg C))\geq 0$$
$$(6)[\text{as,:}\, P(\neg C)\geq 0\ \land 1-x=P(\neg C) \rightarrow [[1-x=P(\neg A)]\geq  0]\,\rightarrow-(1)\times P(\neg C)=-1\times(1-x)\leq -1\times (0)=0$$
(7)$$\rightarrow -1\times(1-x)\leq 0 \rightarrow (x-1)=-1\times(1-x)\leq 0$$
$$(8)\rightarrow (x-1)\leq 0\rightarrow (x-1)+1\leq [[0+1]=1]\rightarrow x\leq 1\rightarrow P(C)\leq 1$$ 
as $PR(C)=x$, 
The same argument applies to $P(\neg C) \leq 1$
So as $$[0\leq[P(C)=x]\leq 1] \land \,  [0\leq [P(\neg C)=1-x]\leq 1]$$
So Unless for some strange reason $\{P(A|C),P(A|\neg C)\}\notin [0,1]$ 
to begin with,then  usually $P(A)\in [0,1]$, as, 
 :
$$\text{and as}:\,[y=\text{Max}(\,P(A| \neg C)],P(A|C)\,),\text{where}[(P(A|C),PR(A|\neg C)\leq 1]\,\leftrightarrow [y \leq 1]\,]$$
As the max is one of the two, and they are both $\leq 1$
These are probabilities and in $[0,1]$ as above  question in any case
$$\text{and as}:[\, [\,x_1= \text{Min}(P(A|\neg C),P(A|C)\,)] \, \text{where,both} [(P(A|C),PR(A|\neg C) \geq 0] \,] \leftrightarrow [x_1 \geq 0]\,\,\,]$$
As the min is one of the two, and they are both $\geq 0$
$$\text{where,as,}\,P(A)=(1-x)\times x_1+ x\times y=x_1+x \times(y-x_1)=y+(1-x)\times[x_1-y] $$
$$\text{where,as,}, 0\leq x, \land [x_1\leq y]\rightarrow (y-x1) \geq 0] \rightarrow x \times(y-x_1)\geq 0 \rightarrow [ x_1+ x \times(y-x_1)] \geq x_1$$
$$\text{where,as,}, [\,[1 \leq x \rightarrow (1-x)\geq 0], \land [x_1\leq y]\,]\rightarrow (x_1-y)\leq 0]\, \land (1-x)\geq 0\rightarrow [(1-x)\times (x_1-y)\leq 0]$$
$$[(1-x)\times(x_1-y)]\leq 0 \rightarrow[ y+(1-x)\times(x_1-y) \leq [0+y=y\,]\,] \rightarrow [PR(A)\leq y]\,\,]$$
$$\,y \geq PR(A)\geq x_1$$
$$ \rightarrow P(A) \in [x_1,y]$$:
$$\text{s.t}\,\{[x_1,y] \subseteq[0,1]\,\}\,\leftrightarrow\,(x_1,y) \in [0,1]\,$$
$$x_1 \leq P(A)\leq y \leftrightarrow  0 \leq P(A)\leq 1$$
A: There are some cases where non con-glomerablity arguably fails though as to whether it leads to

*

*$P(C)=0, \,\text{or} P(\neg c)=0$
2.Or when $P(C)\, \text{or} P(\neg C)$ is non measurable.


*Or when there is an issue with belief update/versus revision or subjunctive versus indicative probabilities where the truth/or probabilty of the antecedent $C$ or $ \neg C $ alters the corresponding conditional probabilities.
ie $P(A|C)=x$ iff only if $C$ is false.
For example
IF '$C$ is false' then, given what is true now in actuality,of the evidence, the probability of $A$ given $C$, is $0.6$, however, were $C$ to be actually true, then this evidence would have been different/changed so that the probability of $A$ and thus $P(A|C)$ would be $=0.5$ instead.
This is related to self location  and diachonric belief update,  issues as well. Where, whatever it is which makes the antecedent true, or the antecedent's very truth, changes something else, so as to ensure that said antecedent$C$ entails the the consequent$A$ to different degree, then the antecedent,$C$, actually entails the consequent $A$, when taken in isolation given what is true now. (where $C$ has not yet come about or is false, but is what one would infer about the information $C$ would provide,keeping everything else, as close to actual situation as opposition (with relata to categorical properties for example),
This being  opposed(to the subjunctive sense)emphasized text to what would have change to make $C$, where other things may have to alter, given the laws of nature and other intensional facts)
These often only come apart if at all when the antecedent is false.
Indicative',which is more evidential in kind, based on what is known know as opposed to how things would have changed, given a change in the world.
Example:
If Oswald did not shoot Kennedy, with high probability, someone else did" (as Kennedy was killed in actuality (as a result of Oswald's shooting him presumably)
and so if Oswald did not do it, someone else did, lest Kennedy would not have been killed , and we know he was killed, and that cannot occur unless someone did away with him, so if it was not Oswald it must have been someone else $\text{vis a vis, }$ given his shooting an death, in actuality) etc).
Subjunctive: 'Were Oswald to have not shot Kennedy, then with very low probability,, someone else would have."
"as there were no backup shooters,(presumably) then most likely Kennedy would never have been shot nor killed to begin with, and he so its he cannot have been shot and killed by someone else in some counter-factual scenario if he was never killed to begin with and thus not shot or killed by anyone at all.
then the probability that I would sleep, if well, is 0.5
