# How to solve this kind of Conditional Probability problems?

I have a series of problems that involve events that happen one after another, and where the usual formulas like P(A/B)=P(AB)/P(A) don't seem to help.

Example:

In a horse race there are 3 horses. E12 means horse 1 finishes before horse 2. E123 means horse 1 finishes before 2, which finishes before 3, etc.

If P(E13)=2/3 and P(E23)=1/2 then P(E123)=?

How can I go about solving this kind of problems?

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This is not a conditional probability question. Your particular example has insufficient information to solve it. Here $E123$ is the event of the horses finishing in the order 1, 2, 3. So (assuming the horses can't tie for a position) $$P(E13)=P(E132)+P(E123)+P(E213),$$ $$P(E23)=P(E123)+P(E213)+P(E231)$$ and of course $$1=P(E123)+P(E132)+P(E213)+P(E231)+P(E312)+P(E321).$$ Knowing $P(E13)$ and $P(E23)$ isn't enough to determine any one of the $P(Eijk)$s.