91Ó°ÊÓ

Let the events be defined as follows: - \(E:\) The event that we are interested in (event \(E\) occurs) - \(F:\) The event that stops the process (event \(F\) occurs) Since \(E\) and \(F\) are mutually exclusive, this means that \(P(E \cap F) = 0\). The probability of the complement of \(E\) and \(F\) happening in each trial is given by \(P(E^{c} \cap F^{c}) = 1 - P(E) - P(F)\), since \(P(E \cup F) = P(E) + P(F)\).

Let's denote the probability of event \(E\) occurring before event \(F\) by \(P(E \text{ before } F)\). We will condition this probability on the first trial. There are three scenarios for the first trial: 1. In the first trial, event \(E\) occurs - In this case, the sequence starts with \(E\), and thus, \(E\) occurs before \(F\). The probability of this happening is \(P(E)\). 2. In the first trial, event \(F\) occurs - This means that \(F\) occurs before \(E\), contrary to our desired outcome. The probability of this happening is \(P(F)\). 3. In the first trial, neither event \(E\) nor event \(F\) occurs - This means that the first trial results in the complement of both events, and we are still waiting for the first occurrence of \(E\) or \(F\). The probability of this happening is \(P(E^{c} \cap F^{c}) = 1 - P(E) - P(F)\). Now, if the first trial falls into the third scenario, the process will start over in the next trial as if it were the first trial because all trials are independent. Therefore, we have: \( P(E \text{ before } F) = P(E) + (1-P(E)-P(F)) \cdot P(E \text{ before } F) \)

Now, we want to solve for \(P(E \text{ before } F)\) in the equation: \( P(E \text{ before } F) = P(E) + (1-P(E)-P(F)) \cdot P(E \text{ before } F) \) Rearranging the terms, we get: \( P(E \text{ before } F) - (1-P(E)-P(F)) \cdot P(E \text{ before } F) = P(E) \) Factoring out \(P(E \text{ before } F)\), we have: \( P(E \text{ before } F) (1 - (1-P(E)-P(F))) = P(E) \) Simplifying the equation, we find: \( P(E \text{ before } F) \cdot (P(E) + P(F)) = P(E) \) Lastly, divide both sides by \((P(E) + P(F))\): \( P(E \text{ before } F) = \frac{P(E)}{P(E) + P(F)} \) This demonstrates that the probability of event \(E\) occurring before event \(F\) is given by the formula \(P(E) / [P(E) + P(F)]\).