Poisson process - probability of an arrival in the time interval Customers arrive at a service center according to a Poisson process with a mean interarrival time of 15 minutes. 
1) If two customers were observed to have arrived in the first hour, what is the probability that both arrived in the last 10 minutes of that hour?
2) If two customers were observed to have arrived in the first hour, what is the probability that at least one arrived in the last 10 minutes of that hour?

My understanding: The arrivals are uniformly distributed. Let X=customer arrives in the last 10 minutes. Then $P(X)= \frac 16$. The arrivals are independent, so the arrival of 2 customers would have the probability of $\frac 16*\frac 16=\frac 1{36}$ (answer to 1).
Regarding the 2). Let Y=number of customers arriving in the last 10 minutes. Then $P(Y>=1)=P(Y=1)+P(Y=2)=\frac 16 + \frac 1{36}=0.19$.
Is it correct ?
 A: Your reasoning is correct as well as your answer for 1) but not for 2), due to an incorrect value for $\mathbb P(Y=1)$. Perhaps a simpler way of solving 2) is to calculate the complementary probability:
$$
\mathbb P(Y=0)=\frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36},\qquad \mathbb P(Y\geq 1)=1-\mathbb P(Y=0)=\frac{11}{36}.
$$
Now comparing with your method, we can see that $\mathbb P(Y=1)$ does not equal $\frac16$ as you wrote, but rather $\frac{10}{36}$. How to see it directly? Call the arrival times $T_1$ and $T_2$. Then either $T_1$ falls in the last 10 minutes and $T_2$ falls in the first 50 minutes, or vice versa. Thus
$$
\mathbb P(Y=1)=\frac{1}{6}\cdot \frac56 + \frac56\cdot\frac16=\frac{10}{36}.
$$
A: $\lambda = \frac{1}{15}$
Probability that two customers arrive in the first hour
$$=\frac{e^{-\frac{60}{15}}4^2}{2!}\tag 1$$
Probability that two customers arrive in the last 10 minutes and that 0 customers arrive in the first 50 minutes.  Use $\lambda t$ for the mean of the poisson probability.
$$=\frac{e^{-\frac{10}{15}}(\frac{2}{3})^2}{2!}.\frac{e^{-\frac{50}{15}}(\frac{10}{3})^0}{0!}\tag 2$$
The required probability in part I $$=\frac{(2)}{(1)} =0.0278 $$
Part II 
Probability that atleast one customer arrive in the last 10 minutes has two cases.  Case 1 is one arrive in the first 50 minutes and 1 arrive in the last 10 minutes and Case 2 is none arrive in the first 50 minutes and 2 arrive in the last 10 minutes.
$$\frac{e^{-\frac{50}{15}}(\frac{10}{3})^0}{0!}.\frac{e^{-\frac{10}{15}}(\frac{2}{3})^2}{2!}+\frac{e^{-\frac{50}{15}}(\frac{10}{3})^1}{1!}.\frac{e^{-\frac{10}{15}}(\frac{2}{3})^1}{1!}\tag 3$$
The required probability in Part II $$ = \frac{(3)}{(1)} =0.30556$$
