I am trying to do the following problem:"Rock tickets are sold at a ticket counter. Females and males arrive at times of independent Poisson processes with rates 30 and 20 customers per hour. If exactly two customers arrived in the the first five minutes, what is the probability that both arrived in the first 3 minutes?"
Below is my attempt at a solution, but I am not sure if 0.36 is the correct answer. I am trying to use the definition of conditional probability and independent increments to compute the answer. Could someone perhaps check if my method is sound/correct?
$$P\left(N\left(\frac{1}{20}\right)=2|N\left(\frac{1}{12}\right)=2\right)=\frac{P\left(N(\frac{1}{3})=2\cap N(\frac{1}{12})=2\right)}{P\left(N\left(\frac{1}{12}\right)=2\right)}=\frac{P\left(N(\frac{1}{3})=2\cap N(\frac{1}{12})-N(\frac{1}{20})=2\right)}{P\left(N\left(\frac{1}{12}\right)=2\right)}$$ $$\frac{e^{-50/20}(50/20)^2*e^{-50/30}(50/30)^0}{2!*0!}*\frac{2!}{e^{-50/12}(50/12)^2}$$ $$(50/20)^2*\frac{1}{(50/12)^2}=0.36$$
The first step is to note that the question does not ask about the gender of the customers who arrived, so we should consider the aggregate process. To this end, the process for females is Poisson with intensity $\lambda_1 = 1/2$ per minute and the process for males has intensity $\lambda_2 = 1/3$ per minute. So the aggregate process for both genders has intensity $\lambda = \lambda_1 + \lambda_2 = 5/6$ per minute.
Given that we observed exactly two customers arriving in the first $5$ minute interval, that is to say, $X(5) = 2$ where $X(t)$ is the number of arrivals in $t$ minutes, then we want to calculate $\Pr[X(3) = 2 \mid X(5) = 2]$. This is
$$\Pr[X(3) = 2 \mid X(5) = 2] = \frac{\Pr[(X(3) = 2) \cap (X(5) = 2)]}{\Pr[X(5) = 2]} = \frac{\Pr[X(3) = 2]\Pr[X(2) = 0]}{\Pr[X(5) = 2]}.$$
The reason why the joint probability in the numerator equals the product of the probabilities that there are exactly $2$ arrivals in the first three minutes and $0$ arrivals in the first two minutes, is because the only way the joint condition is satisfied is if there are exactly $2$ arrivals in the first $3$-minute increment, and $0$ arrivals in the next $2$-minute increment. But non-overlapping increments are independent, and increments are stationary for a homogeneous process, thus the $2$-minute increment from time $t = 3$ to $t = 5$ behaves the same way as an increment from $t = 0$ to $t = 2$.
All that now remains is to compute these probabilities via the formula
$$\Pr[X(t) = x] = e^{-\lambda t} \frac{(\lambda t)^x}{x!},$$
which gives us
$$\Pr[X(3) = 2] = e^{-(5/6)3} \frac{((5/6)3)^2}{2!}, \\ \Pr[X(2) = 0] = e^{-(5/6)2} \frac{((5/6)2)^0}{0!}, \\ \Pr[X(5) = 2] = e^{-(5/6)5} \frac{((5/6)5)^2}{2!}.$$
Therefore the desired conditional probability is
$$\Pr[X(3) = 2 \mid X(5) = 2] = \frac{e^{-(5/6)3} e^{-(5/6)2}}{e^{-(5/6)5}} \cdot \frac{((5/6)3)^2/2!}{((5/6)5)^2/2!} = \frac{3^2}{5^2} = \frac{9}{25} = 0.36.$$
So your answer is correct. The only difference between your solution and mine is that I worked in minutes, rather than hours.
That said, the value that we obtained from this computation seems to suggest that there may be a simpler way to reason. Indeed, the homogeneous Poisson process has the property that, conditioned on a fixed number of arrivals in a given time period, the distribution of those arrivals in that interval is uniformly distributed. So in this problem, once we know that exactly two customers arrived in the first five minutes, their arrival times in that five-minute interval are independent and uniformly distributed in that interval. The proof of this fact is something I leave as an exercise for the reader.
With this knowledge, we can immediately compute the desired probability, since the probability that a given customer (of the two that arrived) also arrived within the first $3$ minutes is just $3/5$, and the two customers' arrivals are independent, the probability that both arrive in the first $3$ minutes is simply $(3/5)^2$.
We can also generalize: the probability that exactly one customer arrived in the first three minutes is $$\Pr[X(3) = 1 \mid X(5) = 2] = \binom{2}{1} (3/5)^1 (2/5)^1,$$ since we count an arrival occurring in the first $3$ minutes as a single Bernoulli trial with success probability $p = 3/5$. And more generally, if we observed $X(t) = n$ arrivals in the first $t$ minutes, the probability that exactly $m$ of these arrivals occurred in the first $u$ minutes is $$\Pr[X(u) = m \mid X(t) = n] = \binom{n}{m} (u/t)^m (1 - u/t)^{n-m},$$ when $0 \le u \le t$ and $0 \le m \le n$.