I know that for mutually exclusive events:
$P(A\cup B) = P(A) + P(B) $
However, I'm having some trouble deriving this from the Bayes Theorem for practical cases. Let's say that we have a toy, and two events can happen to that toy that renders it unusable. These events are $A$ and $B$, and their probabilities by themselves are:
$$P(A) = 0.3 \qquad P(B) = 0.1$$
There are three possible combinations of events that can render the toy unusable. These being:
- E1: Event A and not Event B ($A\cap B')$
- E2: Event B and not Event A ($B\cap A')$
- E3: Event A and Event B ($A\cap B)$, which will not happen because since they are mutually exclusive which I guess means $P(A|B) = P(B|A) = 0$
The thing is. I know that the result has to be $P(unusable) = P(E1) + P(E2) + P(E3)$ and that $P(E3) = 0$. However there is something strange I come up with when I try to obtain $P(E1)$ and $P(E2)$.
Having Bayes Theorem $P(A\cap B) = P(B)P(A|B)$:
If I try to get the first event, this being $P(E1) = P(A\cap B')$, I have the following results:
$$P(E1) = P(B')P(A|B')$$
where I make the assumption (I think correct) that
$$P(A|B') + P(A'|B') = 1$$
Therefore, since they are mutually exclusive, I'm guessing that $P(A|B') =
P(A)$. After all, if B has not happened, the probability that A happens should be the probability of A happening (I'm pretty sure somewhere here is the error, but I don't get it).And therefore $P(A'|B') = 1 – P(A)$, which would fulfill the equation.
This gives a result of $P(E1) = 0.9\times 0.3 = 0.27$.
I know this is not the correct answer, because it does not even fulfill the requirement of the intersection being conmutative
If I define $E1^* = (B'\cap A)$ and apply Bayes, I get to:
$$ P(B'\cap A) = P(A)P(B'|A) $$
Where now $P(B'|A) + P(B|A) = 1$, and since they are mutually exclusive and $P(B|A) = 0$, then $P(B'|A) = 1$. Now the probability of the event is
$$P(E1^*) = 0.3\times 1 = 0.3$$
And this is the correct answer, because then $P(E2) = P(B) = 0.1$ and $P(E3) = 0$ and $P(A\cup B) = 0.4$ which is the objective.
Where did I go wrong in the assumptions for the first $E1$ probability? Is not $P(A|B') = P(A)$? If not, why?
Thank you very much for your time.
Best Answer
You mixed "mutually exclusive" and "stochastical independent"
$$P(A|B)=P(A)$$ means that $A$ and $B$ are stochastically independent. Another way to formulate this is $$P(A\cap B)=P(A)\cdot P(B)$$ This has nothing to do with the formula $$P(A\cup B)=P(A)+P(B)$$ which is true when $A$ and $B$ cannot happen simultaneously.