Given a box with 100 balls with colors red and blue and 2 Poisson random variables such that $X ~ Poi(n)$ and $Y ~ Poi(2n)$ such that $X+1$ is the number of blue balls and $Y+1$ is the number of red balls in the box.
A person starts to take out ball by ball from the box without putting it back, What is the probability that the last ball is blue?
I do have a sense that is something like ${X+1}/{100}$, and out of the total balls, X is around $1/3$ from the total balls in the box, further more i understand that it's irrelevant if it's last ball or first ball so it makes sense that it suppose to be the number of blue balls divided by the total balls.
so it might be something like $33/100$ but i don't feel comfortable with this answer so i hope to get some help!
If I understand correctly, what you mean is that the numbers of red and blue balls were drawn from these Poisson distributions (plus one), and the sum turned out to be $100$.
If so, we can take out the two $+1$ balls (one blue, one red) and are left with two Poisson variables that sum to $98$. Conditional on their sum, these variables are binomially distributed; each ball is independently blue with probability $\frac n{n+2n}=\frac13$ and red with probability $\frac{2n}{n+2n}=\frac23$. Thus the expected number of blue balls is $\frac{98}3$. Add back the blue $+1$ ball, and we expect to have $\frac{98}3+1=\frac{101}3$ blue balls, so the probability that any given (e.g. the last) ball is blue is $\frac{\frac{101}3}{100}=\frac{101}{300}$.