Jamie rolls a die multiple times. find the probability that she rolls her first 5 before she rolls her second even number Jamie rolls her fair 6-sided die multiple times. 
Find the probability that she rolls her first 5 before she rolls her second (not necessarily distinct) even number?
This is what I have so far...
the probability that she rolls a 5 is $\frac{1}{6}$
the probability that she rolls an even number if $\frac{1}{2}$
Now I'm stuck...can anyone help me get to the final product? please be clear and concise.

Consider the following two events:


*

*$A:=\left\{\mbox{sequence of rolls containing one or no even number, and ending in a 5}\right\}$

*$B:=\left\{\mbox{sequence of rolls containing some even number}\right\}$


The probability of interest, in a sequence of independent dice rolls, is
$$
P\left(A \mbox{ followed by } B\right){}={}P\left(B\, |\, A\right)P\left(A\right)\,.
$$
Note that, because the rolls are independent and the event $B$ is an almost-sure event (that is, the probability of it not occurring is zero), we have
$$
P\left( B \, | \, A\right){}={}P\left( B \right){}={} 1\,.
$$
So, 
$$
P\left(A \mbox{ followed by } B\right){}={}P\left(A\right)\,.
$$
We proceed, therefore, to compute the event $A$ as follows (making liberal use of independence):
$$
\begin{eqnarray*}
P\left(A\right)&{}={}&\sum\limits_{k=1}^{\infty} \left(P\left(k\mbox{ rolls, no even, ending in }5\right) {}+{} P\left(k\mbox{ rolls, one even, ending in }5\right)\right) \newline
&{}={}&\sum\limits_{k=1}^{\infty} \left(\frac{1}{6}\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{6}{k-1\choose 1}\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline
&{}={}&\frac{1}{6}\sum\limits_{k=1}^{\infty} \left(\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{2}\left(k-1\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline
&{}={}&\frac{1}{6}\left(\frac{1}{\left(1-\frac{1}{3}\right)}{}+{}\frac{1}{2\left(1-\frac{1}{3}\right)^2}\right)\newline
&{}={}&\frac{7}{16}\,.
\end{eqnarray*}
$$   
