Conditional Probability of Two Poisson RV's

Question

Question:

During a given year for a circus performer, let X represent the number of minor accidents, and let Y represent the number of major accidents.

The joint distribution is:

$f(x,y) = \Large \frac{e^{-1}(.2)^y(.8)^x}{x!y!}$

What is the probability that the performer has at most 1 major accident and at most 2 two total accidents within a year.

Attempt:

I have done this problem several different ways, but my thought was to calculate

$\small P(X=0,Y=0) + P(X=1,Y=0) + P(X=0,Y=1) + P(X=1,Y=1) + P(X=2,Y=0)$.

That would result in:

$e^{-1}(1 + .8 + .2 + .16 + .32) = 2.48e^{-1} = .91234$

Is this correct?

Answer 1

You should write $P[X=0,Y=0]$ and so on, because you are adding probabilities. Otherwise it looks fine to me.

Answer 2

You want to compute $P\left(\{X\le 2 , Y\le 1\}\right)$ while $X+Y\le 2$, which is $P\left(\cup_{x=0}^2\{Y=0, X=x\} \right) + P\left(\cup_{x=0}^1\{Y=1, X=x\} \right)$ Note that $X$ and $Y$ are independent random variables, such that $X\sim P(0.8)$ and $Y\sim P(0.2)$, hence, \begin{align} &P\left(\cup_{x=0}^2\{Y=0, X=x\} \right) + P\left(\cup_{x=0}^1\{Y=1, X=x\} \right)\\ &=P(Y=0)\sum_{x=0}^2P(X=x)+P(Y=1)\sum_{x=0}^1P(X=x)\\ &=e^{-0.2}\sum_{x\in \{0,1,2\}}\frac{e^{-0.8}0.8^x}{x!} + 0.2e^{-0.2}\sum_{x\in \{0,1\}}\frac{e^{-0.8}0.8^x}{x!} \end{align}