I am given the following question:
Suppose that the times it takes for two students to solve a certain homework problem are independently and identically distributed according to the distribution $Poiss(\lambda)$.
Find the probability that one of the students will take at least twice as long as the other one to solve the problem.
What I did: Since $X,Y$ are independent $P_{Y|X}(y|x)=P(Y=y|X=x)=P(Y=y)$
Given some value, $k$, of $X$: The probability that it takes the second student at least twice as long to do the homework is $P(Y\geq2k)$.
Hence the probability that it takes the second student at least twice as long to do the homework is, according to Law of total probability, $\sum_{k=1}^{\infty}P(X=k)P_{Y|X}(Y\geq2k|X=k)$
$=\sum_{k=1}^{\infty}P(X=k)\cdot P(Y\geq2k)$
$=\sum_{k=1}^{\infty}P(X=k)\cdot(1-P(Y<2k))$
$=\sum_{k=1}^{\infty}P(X=k)\cdot(1-\sum_{j=1}^{2k-1}P(Y=j))$
$=\sum_{k=1}^{\infty}e^{-\lambda}\frac{\lambda^{k}}{k!}\cdot(1-\sum_{j=1}^{2k-1}e^{-\lambda}\frac{\lambda^{j}}{j!}))$
and this is where I am stuck.
Can someone please help me continue on calculating this sum, or maybe suggest a different approach ?