I am trying to solve a previous exam question for one of my courses, but there are no solutions.I've tried to somehow link this to the uniform distribution since we know exactly how many tourists arrive, but I can't get an answer - could someone please provide a solution. Thanks
Let $A$ be the event exactly one arrived in the first $10$ minutes, and $B$ the event $20$ arrived in the first hour. We want $\Pr(A\mid B)$, which is $\Pr(A\cap B)/\Pr(B)$.
Easily, $\Pr(B)=e^{-\lambda}\frac{\lambda^{20}}{20!}$.
To calculate $\Pr(A\cap B)$, multiply $\Pr(A)$ by the probability $19$ people arrived in the next $5/6$ of an hour. We have $$\Pr(A)=e^{-\lambda/6}\frac{(\lambda/6)^1}{1!},$$ and the probability of $19$ arrivals in the next $5/6$ of an hour is $$e^{-5\lambda/6}\frac{(5\lambda/6)^{19}}{19!}.$$ Multiply. We get $e^{-\lambda}\frac{\lambda^{20}(1/6)(5/6)^{19}}{19!}$.
Divide by $\Pr(B)$. There is a fair bit of cancellation, and we end up with $20(1/6)(5/6)^{19}$.
Remark: Note that our probability has a familiar "binomial distribution" shape.