Poisson probability with Bayes? The number of goals scored every month is a Poisson with lambda 5 : 
$$
P(X=x) = \frac{e^{-5}5^x}{x!} (x=0,1,2,3,4....)
$$
What is the probability of at least 4 goals scored next month given two goals scored next month.
I need to compute the following : 
P(x >= 4 | X=2)
$$
P(x \ge 4)=1 - \sum_{i=0}^3 \frac{e^{-5}5^i}{i!}
$$
$$
P(x = 2)=\frac{e^{-5}5^2}{2!}
$$
How can these probabilities be combined to produce the probability of at least 4 goals scored next month given two goals scored next month ?
This seems like a Bayes theorom problem as it's a conditional probability but I'm unsure how to map the provided information into the formula ?

The problem statement isn't quite clear but it looks like it wants you to find $\mathsf P(X\geq4\mid X\geq 2)$
It's clear then from Bayes' Theorem that
$$\mathsf P(X\geq4\mid X\geq 2)=\frac{\mathsf P(X\geq4,X\geq 2)}{\mathsf P(X\geq2)}=\frac{\mathsf P(X\geq4)}{\mathsf P(X\geq2)}=\frac{1-\mathsf P(X\leq3)}{1-\mathsf P(X\leq1)}$$
In R we get
$$\mathsf P(X\geq4\mid X\geq 2)\approx 0.766$$
> (1-ppois(3,5))/((1-ppois(1,5)))
[1] 0.7659392

A simulation in R agrees with this result
x<-rpois(10^6,5)
x<-subset(x,x>=2)
mean(x>=4)

0.7659072

