In a home work problem $E$ and $F$ are mutually exclusive events in the sample space of an experiment. The experiment is repeated until either event $E$ or event $F$ occurs. Show that the probability that event $E$ occurs before event $F$ is given by \begin{equation} \frac{P(E)}{P(E) + P(F)}. \end{equation}

I am assuming that this is a geometric processes with prob. that $E$ occurring is $p$ and $F$ occurring is $q$. Then \begin{equation} P(E) = (1-(p + q))^{n-1} p \end{equation} \begin{equation} P(F) = (1 - (p + q))^{N-1}q \end{equation} where $n$ it the number of trials for event $E$ to occur and $N$ is number of trials for $F$ to occur.

Is this approach correct. I am not able to proceed from here. Can this processes be anything other than a geometric processes.


It is more easy to look at it as a conditional probability.

The main question is: what is the probability that $E$ occurs under the condition that one of the events occur?

Something like: $$P\left(E|E\cup F\right)=\frac{P\left(E\cap\left(E\cup F\right)\right)}{P\left(E\cup F\right)}=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$$

Every time that none of the events occurs is irrelevant.

  • $\begingroup$ Thank you for your answer. I had originally considered this as a conditional probability problem, the mistake that I had done was computing P(E|F), so that left me with P(E). This makes more sense now! $\endgroup$ – ashtrix69 Feb 1 '14 at 20:25
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Feb 1 '14 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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