Airport travellers arive by a Poisson proces with rate $1000$ per hour. Of those travellers, $10\%$ is superficially inspected and an additional $1\%$ is extremely careful inspected. That is $11\%$ is inspected. Given that in $10$ minutes $15$ people have been superficially inspected, what is the probability that exactly $4$ people are extremely careful inspected?
If conditioning on the number of people (say $N$) arriving and summing over all possible values and the probability of $N$ people arriving, I can come up with: $\mathbb{P}(4\ $ people thoroughly inspected$)=\sum_x\mathbb{P}(4\ $ people thoroughly inspected$ \mid x\ $people arrive in $10$ minutes$)\mathbb{P}(x\ $people arrive in $10$ minutes$)=$ $$\sum_{x\geq 4} \binom {x}{4} \frac{1}{100}^4 \frac{99}{100}^{x-4} \mathrm{e}^{-\frac{1000}{6}} \frac{\frac{1000}{6}^{x}}{x!}$$ But then I do not use the given information, could someone help?