showing independence of Poisson distribution

Question

I was doing some past exams on distributions, and I can not do part e), nor do I understand the mark scheme. The first 4 questions I did as workings below, but can anyone help me try to understand and show part e)?

Answer 1

We suppose Archie's emails per day $(X_1,X_2)$ are distributed as a Poisson($\lambda$), so his email count over two days $(Y = X_1+X_2)$ is distributed as Poisson($2\lambda$). The probability that Archie received exactly 10 emails on each of the first two days, given that he received a total of 20 emails over the two days, is $$P = \frac{P(X_1=10) P(X_2=10)}{P(Y=20)} = \frac{\left(\frac{\lambda^{10} e^{-\lambda}}{10!}\right)^2}{\frac{(2\lambda)^{20} e^{-2\lambda}}{20!}} = \frac{20!}{(10!)^2 2^{20}}$$ which does not depend on $\lambda$.