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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.

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  • $\begingroup$ $P(A\cap B')=P(A)$ since the events $A,B$ are mutually exclusive. $\endgroup$ – lulu Mar 1 at 12:07
  • $\begingroup$ Yeah that's the result from $P(E1^*)$, however I didn't get to that when doing the conmutative of the intersection, and I'm not able to see the mistake $\endgroup$ – Daniel V. Mar 1 at 12:20
  • $\begingroup$ It is not true that $P(A\,|\,B')=P(A)$. Since $A$ lives entirely in $B'$, knowing that $B'$ occurred is evidence for $A$. As a simpler instance, suppose that you are throwing a fair die and that $A$ is the event "you throw a $6$" and $B$ is the event "you throw an odd value". Then $A$ and $B$ are mutually exclusive and $P(A\,|\,B')$ is $\frac 13$ not $\frac 16$. $\endgroup$ – lulu Mar 1 at 12:35
  • $\begingroup$ I see, so, $P(A|B')$ would have to be equal to $\frac{P(A)}{P(B')}$? $\endgroup$ – Daniel V. Mar 1 at 15:07
  • $\begingroup$ Yes, that is correct and it follows from the definition of conditional probability (and the clear fact that $P(A\cap B')=P(A)$). $\endgroup$ – lulu Mar 1 at 15:41
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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.

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  • $\begingroup$ Sorry, I don't follow :(. I'm not saying $P(A|B) = P(A)$, I mean that these two are mutually exclusive, so $P(A|B) = 0$. What I got to was that $P(A|B') = P(A)$, which I'm not sure is right. I defined the events as mutually exclusive, not independent. I cannot do $P(A\cap B) = P(A)·P(B)$ since that is only for independent events, and these are dependent events. $\endgroup$ – Daniel V. Mar 1 at 12:18
  • $\begingroup$ Didn't you conclude $P(A|B')=P(A)$ from $A$ and $B'$ being mutually exclusive ? If not, I misunderstood the question. $\endgroup$ – Peter Mar 1 at 12:21
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    $\begingroup$ I concluded it a bit from $P(A|B') + P(A'|B') = 1$, and then $P(A|B') = P(A)$ and $P(A'|B') = 1 - P(A)$, but I see now that that is a mistake, thank you! $\endgroup$ – Daniel V. Mar 1 at 15:09

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