The independent random variables $X$ and $Y$ have Poisson distribution with mean $2$ and $3$ respectively. Given that $X+Y=5$, find the probability that $X=1$ and $Y=4$.
Given $X+Y=5$, find probability that $X=1$ and $Y=4$. $X$ and $Y$ follows Poisson distribution with mean $2$ and $3$ respectively.
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discrete-mathematics
1 Answers
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By convolution formula of independent Poisson random variables, the sum is Poisson with parameter being the sum of the parameters:
$X+Y\sim\text{Poisson}(5)$, so $P(X+Y=5)=\frac{5^5e^{-5}}{5!}$
Then: $$P(X=1,Y=4|X+Y=5)=\frac{P(X=1,Y=4)}{P(X+Y=5)}=\frac{P(X=1)P(Y=4)}{p(X+Y=5)}=\frac{\frac{2^1e^{-2}}{1!}\frac{3^4e^{-3}}{4!}}{\frac{5^5e^{-5}}{5!}}=2\left(\frac{3}{5}\right)^4$$
EDIT: One may use another result (if only to verify), namely if $X\sim\text{Poisson}(\lambda)$, and $Y\sim\text{Poisson}(\mu)$, then $Z=X|X+Y=n\sim\text{Binomial}\left(n,\frac{\lambda}{\lambda+\mu}\right)$
In this case, one has $Z\sim\text{Binomial}\left(5,\frac{2}{5}\right)$, and:
$$P(X=1,Y=4|X+Y=5)=P(Z=1)={5\choose 1}\left(\frac{2}{5}\right)^1\left(\frac{3}{5}\right)^4$$