This is the question I am working on:

For two events, M and N, P(M) = 0.4, P(N|M) = 0.6, and P(N|M') = 0.4. Find P(M|N)

I am stuck on this part of the question:

Which branch of the tree diagram shows the probability that both M and N occur?

Tree Diagram here

The answer to this one is 1.

I have no clue. I have looked up Bayes' Theorem and have not found any questions similar to this on google.

Sorry if this is too simple but I can't for the life of me figure this out.

  • $\begingroup$ It is $P(M \cap N)=P(M)\cdot P(N|M)$. $ \ \ $ $P(M \cap N)=0.6\cdot 0.4=0.24$ is the probability that the events $M$ and $N$ occur. Thus the first branch is correct. $\endgroup$ – callculus Jul 2 '16 at 1:02

Notice that the first node, which I colored blue, is the start. Nothing has happened yet. If I follow that branch up to the next node (light blue) , then this represents that $M$ has occurred. Since $M$ has occurred, at this node, we say "given $M$". In other words, it is given that $M$ has occurred. Next, we have two options, follow the branch up or follow it down. If we follow up, then $N$ occurs. If we follows down, then $N$ does not occur.

If we follow up, then $N$ occurs. So all together, $M$ and $N$ occur. So, the answer is $1$. I colored our path in red.

enter image description here


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