Conditioning in a Poisson process. Question. Let $X(t)$ be a Poisson process modelling the arrival of alpha particles at a detector after $t$ hours with a rate of $2$ per hour. Let $Y(t)$ be a Poisson process modelling the arrival of beta particles at a detector after $t$ hours with a rate of $1$ per hour.
Given that exactly $3$ alpha particles and exactly $2$ beta particles arrive during the first hour, what is the probability that more beta particles than alpha particles arrived during the first $30$ minutes?

So, we are required to work out $\Bbb P(Y(1/2)>X(1/2)|X(1)=3, Y(1)=3)$. Now, if we let $(y,x)$ denote the possible combinations of beta to alpha particles arriving in the first half an hour; there are a total of three possible combinations that have beta particles exceeding alpha particles, namely $(2,0), (2,1), (1,0)$.
But then what? I know this result in regards to conditioning:
Let $(X(t): t\geq 0)$ be a Poisson process of rate $\lambda$ with $n\in\Bbb N$. Then, for any $m\in\{0,...,n\}$ $$\Bbb P(X(t)=m|X(s)=n)={n\choose{m}}\Big(\frac{t}{s}\Big)^m\Big(1-\frac{t}{s}\Big)^{n-m}=\Bbb P\Big(\text{Bin}\Big(n,\frac{t}{s}\Big)=m\Big),$$
and feel as though this will play a part in the solution; but I'm not quite sure how to use this to my advantage.

Thanks in advance for any assistance here.

Edit. A very shameless bump as I am still stuck on this problem.
 A: Just for completeness; I will post the answer to this question which I have finally got to:
Let y-x be the possible combinations of beta and alpha particles that can appear at the counter in the first half an hour. Now, there are three possible combinations where the number of beta particles exceed the number of alpha particles; namely 2-0, 2-1 and 1-0.
        Since these events are disjoint, we have $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)>X\Big(\frac{1}{2}\Big)|X(1)=3, Y(1)=3\Big),$$ that is, we take the product of these three probabilities by independence $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)=2, X\Big(\frac{1}{2}\Big)=0|X(1)=3, Y(1)=2\Big),$$ $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)=2, X\Big(\frac{1}{2}\Big)=1|X(1)=3, Y(1)=2\Big),$$ $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)=1, X\Big(\frac{1}{2}\Big)=0|X(1)=3, Y(1)=2\Big).$$ Now, observing that the arrival of alpha and beta particles are independent of each other; we may further decompose these expressions as follows. We have $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)=2,|Y(1)=2\Big)\Bbb P\Big(X\Big(\frac{1}{2}\Big)=0,|X(1)=3\Big),$$ $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)=2,|Y(1)=2\Big)\Bbb P\Big(X\Big(\frac{1}{2}\Big)=1,|X(1)=3\Big),$$ $$\Bbb P\Big(Y\Big(\frac{1}{2}\Big)=1,|Y(1)=2\Big)\Bbb P\Big(X\Big(\frac{1}{2}\Big)=0,|X(1)=3\Big).$$
        Multiple applications of the conditioning theorem gives, in order, the probabilities $$\frac{1}{4},\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2}, \frac{1}{8}.$$ So, we have $$\frac{1}{4}\cdot\frac{1}{8}+\frac{1}{4}\cdot\frac{3}{8}+\frac{1}{2}\cdot\frac{1}{8}=\frac{1}{32}+\frac{3}{32}+\frac{1}{16}=\frac{6}{32}=\frac{3}{16}.$$
