Question

The independent random variables X and Y have distributions Po(2) and Po(3) respectively. Given that X + Y = 5, find the probability that X = 1 and Y = 4

Answer 1

I have got
P(X=1) x P(Y=4)
but then the answers say to divide by P(X+Y = 5) which I'm not to sure why? Help?

Answer 2

Conditional probability:
\[P(X=1\text{ and }Y=4~|~X+Y=5)\]
which reads as "the probability that \(X=1\) and \(Y=4\) given that \(X+Y=5\). You compute this as follows:
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
which in this case would you compute
\[P(X=1\text{ and }Y=4~|~X+Y=5)=\frac{P\bigg(\big[X=1\text{ and }Y=4\big]\cap\big[X+Y=5\big]\bigg)}{P(X+Y=5)}\]

Answer 3

So for the top part, [X=1 and Y=4] intersection [X+Y = 5] the probability for the X+Y=5 part is 1?