Poisson probability with Bayes?

Question

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 ?

Answer 1

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