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.

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.


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?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer

Your Answer


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.