Working my way through the following problem:
Problem
Suppose that $E$ and $F$ are mutually exclusive events of an
experiment. Show that if independent trials of this experiment are
performed, then $E$ will occur before $F$ with probability
$\frac{ P( E)}{P( E) + P( F)}.$
I have the following come up with the following solution:
Solution
Since
$P( E^c) = P( F)$
Therefore
$\frac{ P( E)}{ P( E) + P( F)} = \frac{ P( E)}{ 1 – P( F) + P( F)} = \frac{ P( E)}{ 1} = P( E)$
But I am unsure if I am able to assume $P( E^c) = P( F)$ as a given?
As well, I am particularly confused by the answer in the solution manual which makes it's argument as follows:
Solution Manual
If $E$ and $F$ are mutually exclusive events in an experiment, then
$P( E \cup F) = P( E) + P( F)$. We desire to compute the probability
that $E$ occurs before $F$ , which we will denote by $p$. To compute
$p$ we condition on the three mutually exclusive events $E$, $F$ , or
$(E \cup F )^c$. This last event are all the outcomes not in $E$ or
$F$. Letting the event $A$ be the event that $E$ occurs before $F$, we
have that$p = P( A|E) P( E) + P( A|F) P(F ) + P( A|(E \cup F )^c) P( (E \cup F )^c)$
$P( A|E) = 1$
$P( A|F) = 0$
$P( A|(E \cup F)^c) = p$
since if neither $E$ or $F$ happen the next experiment will have $E$
before $F$ (and thus event $A$ with probability $p$). Thus we have
that$p = P( E) + p P( (E \cup F)^c)$
$= P( E) + p (1 − P( E \cup F))$
$= P( E) + p (1 − P( E) − P( F))$
Solving for $p$ gives
$p = \frac{ P( E)}{ P( E) + P( F)}$
as we were to show.
Specifically his statement
since if neither $E$ or $F$ happen the next experiment will have $E$
before $F$ (and thus event $A$ with probability $p$)
Best Answer
To determine the probability that $E$ occurs before $F$, we can ignore all the (independent) trials on which neither $E$ nor $F$ occurred, that is, $(E\cup F)^c$ occurred, since we are going to repeat the experiment until one of $E$ and $F$ does occur. So, look at the trial of the experiment on which one of $E$ and $F$ has occurred for the very first time. We are given that on this trial, the event $E \cup F$ has occurred. But, we don't yet know which of the two has occurred. So, given the knowledge that $E \cup F$ has occurred, what is the conditional probability that it was $E$ that occurred (and so $E$ occurred before $F$ since this is the first time we have seen either $E$ or $F$)?
$$P(E \mid (E \cup F)) = \frac{P(E(E \cup F))}{P(E \cup F)} = \frac{P(E \cup EF)}{P(E) + P(F) - P(EF)} = \frac{P(E)}{P(E)+P(F)}$$ since $P(EF) = P(\emptyset) = 0$.
Alternatively, let $G = (E\cup F)^c = E^c \cap F^c$ be the event that neither $E$ nor $F$ occurs on a trial of the experiment. Note that $P(G) = 1 - P(E) - P(F)$. Then, the event $E$ occurs before $F$ if and only if one of the following compound events occurs:
$$ E, (G, E), (G, G, E), \ldots, (\underbrace{G, G, \ldots, G,}_{n-1} E), \ldots $$
where $(\underbrace{G, G, \ldots, G,}_{n-1} E)$ means $n-1$ trials on which $G$ occurred and then $E$ occurred on the $n$-th trial. The desired probability is thus
$$P(E ~\text{before}~ F) = P(E) + P(G)P(E) + [P(G)]^2P(E) + \cdots = \frac{P(E)}{1 - P(G)} = \frac{P(E)}{P(E)+P(F)}.$$