Consider X∼Poisson(λ1)X \sim \text{Poisson}(\lambda_1)X∼Poisson(λ1) and Y∼Poisson(λ2)Y \sim \text{Poisson}(\lambda_2)Y∼Poisson(λ2) independent. What is the conditional distribution P(X=k∣X+Y=n)P(X=k | X+Y=n)P(X=k∣X+Y=n)?
Poisson(λ1+λ2)\text{Poisson}(\lambda_1 + \lambda_2)Poisson(λ1+λ2)
Binomial(n,λ1λ1+λ2)\text{Binomial}(n, \frac{\lambda_1}{\lambda_1+\lambda_2})Binomial(n,λ1+λ2λ1)
Geometric(λ1λ1+λ2)\text{Geometric}(\frac{\lambda_1}{\lambda_1+\lambda_2})Geometric(λ1+λ2λ1)
Normal(n,λ1+λ2)\text{Normal}(n, \lambda_1+\lambda_2)Normal(n,λ1+λ2)