I am having trouble finding P(X≤Y|X+Y =1). I know a conditional probability is going to be (P(X≤Y)AND P(X+Y =1)/ P(X+Y =1).
I am unsure how to find these separate variables in the multivariable sense. I realize they are independently distributed, so does that mean you can multiply individual poisson probabilities?
I don't know how to visualize the separate parts of this conditional probability, so I think that is where my main problem is. I understand these problems for conditional distributions, but the discrete problems I haven't visualized well.
Would the denominator be ((e^-1/1!))(3e^-3/1!)) ?
How would you then find the probability X
Since $X+Y=1$ we must have either $(X,Y)=(1,0)$ or $(X,Y)=(0,1)$
We compute:
$$P(1,0)=P(X=1)\times P(Y=0)=\frac {1^1\times e^{-1}}{1!}\times \frac {3^0\times e^{-3}}{0!}=e^{-4}$$
$$P(0,1)=P(X=0)\times P(Y=1)=\frac {1^0\times e^{-1}}{0!}\times \frac {3^1\times e^{-3}}{1!}=3e^{-4}$$
Thus $P(X+Y=1)= 4e^{-4}$
As the only case in which $X≤Y$ and $X+Y=1$ is $(X,Y)=(0,1)$ we get the final answer $$\frac {3e^{-4}}{4e^{-4}}=\boxed {\frac 34}$$