Question

Let \(E\) and \(F\) be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event \(E\) or event \(F\) occurs. What does the sample space of this new super experiment look like? Show that the probability that event \(E\) occurs before event \(F\) is \(P(E) /[P(E)+P(F)]\). Hint: Argue that the probability that the original experiment is performed \(n\) times and \(E\) appears on the \(n\) th time is \(P(E) \times(1-p)^{n-1}, n=1,2, \ldots\), where \(p=P(E)+\) \(P(F)\). Add these probabilities to get the desired answer.

Short answer

In the super experiment where the original experiment is repeated until either E or F occurs, the probability that event E occurs before event F is given by the sum of an infinite geometric series: \(P(E \,\, \text{prior to} \,\, F) = \sum_{n=1}^{\infty} P(E) \times (1-p)^{n-1}\), where \(p=P(E)+P(F)\). Applying the formula for the sum of an infinite geometric series, we get \(P(E \,\, \text{prior to} \,\, F) = \frac{P(E)}{P(E)+P(F)}\), which proves the desired relationship.

Step-by-step solution

Describe the super experiment

In the super experiment, the original experiment is repeated until either E or F occurs, and we define success as the event E occurring before F. The sample space for this super experiment can be formed by concatenating the outcomes of the individual experiments.

Compute the probability E occurs at the n-th attempt

The hint tells us to argue that the probability the original experiment is performed n times and E appears on the n-th time is \(P(E) \times(1-p)^{n-1}\), where \(n=1,2, \ldots\) and \(p=P(E)+P(F)\). Intuitively, this means that E didn't occur in the first \(n-1\) experiments, and then it finally occurs in the n-th attempt. Probability of E not occurring in a specific experiment is \((1-P(E))\). Since E and F are mutually exclusive, the probability that the experiment results in neither E nor F (which implies that the experiment will be repeated) is \((1-p)=(1-[P(E)+P(F)])\). Thus, the probability of E not occurring in the first \(n-1\) experiments and then occurring in the n-th experiment is given by \(P(E) \times (1-p)^{n-1}\).

Compute the probability E occurs before F

Now we have the probability of E occurring at exactly the n-th attempt for all n. The probability that event E occurs before F can be found by summing up all these individual probabilities: \(P(E \,\, \text{prior to} \,\, F) = \sum_{n=1}^{\infty} P(E) \times (1-p)^{n-1}\) This is a geometric series with the first term equal to \(P(E)\) and the common ratio \((1-p)\). The sum of an infinite geometric series can be written as: \(S = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio. Applying this formula, we get: \(P(E \,\, \text{prior to} \,\, F) = \frac{P(E)}{1-(1-p)} = \frac{P(E)}{p}\) Since \(p=P(E)+P(F)\), the probability that event E occurs before event F is: \(P(E \,\, \text{prior to} \,\, F) = \frac{P(E)}{P(E)+P(F)}\) Thus, we have demonstrated that the probability that event E occurs before event F is \(P(E) /[P(E)+P(F)]\).