Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
1  
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

share|improve this answer

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 $B$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.