I’ve managed to get the correct answer to this question but I really doubt if I actually solved it or just stumbled upon the solution.
The Question: a certain fridge has one cookie in it, the number of people opening the fridge during the day is poissonian with parameter $\lambda = 6$. If a person opens the fridge and the cookie is uneaten, he’ll eat it with probability $p=\tfrac23$. What’s the probability 3 people opened the fridge if the cookie is known to be uneaten?
Now I’ve tried solving it with Bayes’ rule but was unable to compute $P(\text{cookie remains uneaten})$ so I abandoned it in favor of something I remembered.
A poissonian distribution affected by an independent event with probability p is distributed as $p_X(k) = e^{-\lambda p } \frac{(\lambda p)^k}{k!}$ so if I choose $p=\tfrac13$ I might be able to say that probability of 3 people coming to the fridge and choosing not to eat the cookie is $$p_X(3) = e^{-1/3 \cdot 6} \cdot \frac{2^3}{3!}= \frac43e^{-2}$$ which is the correct answer.
I’m unsure if my calculation is correct, when is this modification applicable and what’s the justification for this, and if it’s not how am I supposed to approach this question?
I'm not sure about the solution you've arrived at, however if you want a solution utilizing Bayes' theorem, here is one:
Let $\mathsf{N}$ denote the number of people opening the fridge, i.e. $\mathsf{N}\sim \mathrm{Poisson}(6)$ and let $A$ be the event that the cookie remains uneaten. By your description, you've probably found that $$ \mathbb{P}(\mathsf{N}=3\mid A) =\frac{\mathbb{P}(A\mid \mathsf{N}=3)\cdot\mathbb{P}(\mathsf{N}=3)}{\mathbb{P}(A)} =\frac{\frac{1}{3^3}\cdot\frac{6^3\mathrm{e}^{-6}}{3!}}{\mathbb{P}(A)}, $$ and the tricky part is now finding $\mathbb{P}(A)$. One useful trick is using the law of total probability. Indeed, using this we find that $$ \mathbb{P}(A) =\sum_{n=0}^{\infty}\mathbb{P}(A\mid \mathsf{N}=n)\cdot \mathbb{P}(\mathsf{N}=n) =\sum_{n=0}^{\infty}\frac{1}{3^n}\cdot\frac{6^n\mathrm{e}^{-6}}{n!} =\mathrm{e}^{-6}\sum_{n=0}^{\infty}\frac{2^n}{n!} =\mathrm{e}^{-6}\cdot \mathrm{e}^{2}. $$ Plugging this back into the above, we find that $$ \mathbb{P}(\mathsf{N}=3\mid A) =\frac{\frac{1}{3^3}\cdot\frac{6^3\mathrm{e}^{-6}}{3!}}{\mathrm{e}^{-6}\cdot\mathrm{e}^{2}} =\frac{4}{3}\mathrm{e}^{-2}. $$