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Disclaimer: This is homework, however I am not looking for an answer. I'm only trying to understand the actual question.

I'm given four mutually exclusive and exhaustive events: $A$, $B$, $C$, and $D$. I'm also given $P(A)$, $P(B)$, $P(C)$ and $P(D)$. There is also some minor event $M$, for which I have $P(M|A)$, $P(M|B)$, $P(M|C)$ and $P(M|D)$.

Now for the actual question:

Given that a problem is due to the problem $M$, what is the probability that $B$ occurs?

What exactly is this saying? Would it be proper to say that is it asking for:

$$ P(B|M) $$

Which could be solved using:

$$ \frac{P(B)}{P(M)} $$

Seeing as they are mutually exclusive?

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Yes, they are asking for $P(B|M)$. No, $M$ and $B$ are not mutually exclusive. And anyway $P(B|M)$ is not given by $P(B)/P(M)$. –  Fabian Jan 23 '12 at 17:49
And if M and B were mutually exclusive, P(B|M) = 0 by definition anyway... –  Chad Miller Jan 23 '12 at 17:58
@ChadMiller: thanks for clearing that up –  maxmackie Jan 23 '12 at 18:01

2 Answers 2

up vote 2 down vote accepted

Yes it is asking for $P(B|M)$.

No it cannot be solved as $\dfrac{P(B)}{P(M)}$ since $B$ and $M$ are not simply related. For this to work you would need $B$ to be a subevent of $M$.

Instead you should use conditional probability / Bayes' theorem

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The correct conditional probability formula is $$P(B|M)=\frac{P(B\cap M)}{P(M)},\qquad\qquad(\ast)$$ and that is what we should start from. You are essentially assuming that $P(B\cap M)=P(B)$. That can hardly ever be true. Typically your $P(B)/P(M)$ will be greater than $1$, so cannot even be a probability.

Let us compute $P(B\cap M)$. We have $$P(M|B)=\frac{P(B\cap M)}{P(B)}.$$ You have been told $P(M|B)$, and $P(B)$, so you can find $P(B\cap M)$.

The only thing more that we need in order to use $(\ast)$ is $P(M)$. Now $M$ can happen in $4$ disjoint ways. We could have $A$ and $M$, or $B$ and $M$, or $C$ and $M$, or $D$ and $M$.

We already found $P(B\cap M)$. In a similar way, we can find $P(A\cap M)$, $P(C\cap M)$, and $P(D\cap M)$. Add up these $4$ probabilities to find $P(M)$.

Comment: The problem as it is put has a somewhat abstract tone. It might help to tell oneself a story. Four schools (of course named $A$, $B$, $C$, and $D$) have respectively $300$, $400$, $600$, and $700$ students. We know what proportion of students in each are interested in Mathematics. We want to know the probability that if a randomly chosen student from the $2000$ turns out to be interested in Mathematics, then she comes from school $B$.

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