Expressing "Probability that #successes is an even number" mathematically Needing a little help with my probability concept. Here's the question:

An urn contains $10$ red balls, $20$ green balls and $30$ blue balls. Each trial consists of drawing a ball from the urn with replacement. If either red or blue ball is drawn, the trial is called a success. Suppose that $n$ independent trials are performed and let $P_n$ be the probability that the total number of successes that result is an even number. Find $P_n$ and $\lim \limits_{n \to \infty} P_n$.

My Solution:
$$ P (\text{success}) = \frac{\binom{40}{1}}{\binom{60}{1}} = \frac {2}{3} .$$
Then it is a binomial r.v with parameters $(n, \frac 2 3)$.
How do I express the idea "total number of successes that result is an even number" mathematically?
Thanks for looking at my question. 
 A: Let $P_n$ be the probability that after $n$ trials the number of successes is even.
Let $p$ be the probability of success on any one trial.  In our problem,  $p=2/3$, but we might as well generalize a bit.
The number of successes after $n+1$ trials can be even in two ways: (a) After $n$ trials we had an odd number of successes and we got a success on the $(n+1)$-th trial; or (b) After $n$ trials we had an even number of successes, and we had a failure on the $(n+1)$-th trial.  The probability of (a) is $p(1-P_n)$ and the probability of (b) is $(1-p)P_n$. We therefore have the recurrence
$$P_{n+1}=p(1-P_n)+(1-p)P_n=p+(1-2p)P_n.\qquad (\ast)$$
The recurrence $(\ast)$ is linear, and there are general tools for solving such recurrences.  But the recurrence is particularly simple, as is the physical situation, so we will use a trick.
It is intuitively clear that if $p(1-p)\ne 0$ and $n$ is large, then $P_n$ should be close to $1/2$.  Let  $P_n=1/2+y_n$, and substitute in $(\ast)$. There is a lot of cancellation, and we obtain
$$y_{n+1}=y_n(1-2p). \qquad (\ast\ast)$$
Note that $y_0=1/2$, since if there are $0$ trials, for sure there are $0$ successes.  Each time we increment $n$ by $1$, $y_n$ gets multiplied by $1-2p$. So the sequence $(y_n)$ is the geometric sequence 
$y_n=\frac{1}{2}(1-2p)^n$,
and therefore 
$$P_n=\frac{1}{2}(1+(1-2p)^n).$$
If $p=0$ or $p=1$, $P_n$ is completely determined by the parity of $n$. Suppose now that $p\ne 0$ and $p\ne 1$. Then $|1-2p|<1$, so 
$(1-2p)^n$ approaches $0$ as $n \to\infty$. Thus $P_n$ indeed has limit $1/2$.
Comments: $1$. The recurrence approach can be used with more complicated problems, such as determining the probability that the number of successes after $n$ trials is a multiple of $3$. 
$2$. One can also use an algebraic approach which is basically a rewording of the solution by Didier Piau. Let $F_n(t)=(tp +(1-p))^n$.  Expand $F_n(t)$ using the Binomial Theorem. Evaluate $F_n(t)$ at $t=1$ and at $t=-1$, add up. Suppose that $k$ is odd. Then the terms $\binom{n}{k}p^k(1-p)^{n-k}$ and  $\binom{n}{k}(-p)^k(1-p)^{n-k}$ cancel. Suppose that $k$ is even. Then the terms $\binom{n}{k}p^k(1-p)^{n-k}$ and  $\binom{n}{k}(-p)^k(1-p)^{n-k}$ are equal. It follows that
$$P_n=\frac{1}{2}(F_n(1)+F_n(-1))=\frac{1}{2}(1^n+(1-2p)^n).$$
$3$. When we let $y_n=1/2+F_n$, the recurrence simplified. Consider the general recurrence $P_{n+1}=a+bP_n$, where $b \ne 1$. Make the substitution $F_n=w+y_n$. We get $y_{n+1}=by_n +a+ bw-w$. By setting $w=a/(1-b)$, we get the recurrence $y_{n+1}=by_n$, which is simple to solve.
A: Since $P_n=P(X_n\ \text{is even})$ with $X_n$ binomial $(n,p)$ and $p=\frac23$, $$P_n=\sum\limits_{k\geqslant0}P(X_n=2k)=\sum\limits_{k\geqslant0}{n\choose 2k}p^{2k}(1-p)^{n-2k}.
$$
From this point, it might help to note that, for every $(x,y)$, one can compute
$$
e_n(x,y)=\sum\limits_{k\geqslant0}{n\choose 2k}x^{2k}y^{n-2k}\quad
\text{and}\quad
o_n(x,y)=\sum\limits_{k\geqslant0}{n\choose 2k+1}x^{2k+1}y^{n-2k-1}
$$
both at the same time by considering $e_n(x,y)+o_n(x,y)$ and $e_n(x,y)-o_n(x,y)$.
Edit Once you computed $P_\infty=\lim\limits_{n\to\infty} P_n$, you might wish to check that $P_\infty=\frac23(1-P_\infty)+\frac13P_\infty$ and to explain why this relation, which allows to compute $P_\infty$ without knowing what the exact formula for $P_n$ is, should hold.
