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 $p$)