I’m trying to apply the probabilistic method on this problem, and I’d like some checking on my solution.
We say a $4$-colouring of the vertices of a $k$-uniform hypergraph is rainbow if every edge has all four colours represented. Prove that all $k$-uniform hypergraphs $H$ with $$e(H) \leq \frac{4^{k-1}}{3^{k}}$$ admit a rainbow $4$-colouring. Here, $e(H)$ denotes the number of edges of $H$.
Proof: Consider a random $4$-colouring of $H$, and let $e$ be any edge in $H$. We have: $$\begin{align}\text{Pr}(e \text{ is not rainbow}) &= 4 \cdot \left( \frac{1}{4} \right)^k = 4^{1-k}\\ \Rightarrow \text{Pr}(H \text{ is not rainbow}) &= \text{Pr}(\text{ at least one edge is not rainbow})\\& \leq 4^{1-k} \cdot e(H) \leq \frac{1}{3^k} < 1\,.\end{align}$$ So there is a $4$-coloring of $H$ that is a rainbow coloring.
I feel like my proof is missing some point, as it doesn’t use much of the fact that “$3^k$” appears in the bound for $e(H)$. But I can’t see how the reasoning doesn’t make sense.
Your solution is almost correct, except this: $$\mathbb{P}(e\text{ is not rainbow})=4\,\left(\frac{\color{red}3}{4}\right)^k=\frac{{\color{red}{3^k}}}{4^{k-1}}\,.$$ Therefore, $$\mathbb{P}(H\text{ is not rainbow})\leq e(H)\cdot\frac{{\color{red}{3^k}}}{4^{k-1}}\,{\color{red}\leq} \,1\,.$$ It takes a bit more to conclude that $$\mathbb{P}(H\text{ is not rainbow})\,\color{blue}{<}\,1\,.\tag{*}$$
If there are at least two edges in $H$, say, $e$ and $e'$, then $$\mathbb{P}(e\text{ and }e'\text{ are not rainbow})>0\,.$$ Therefore, $$\mathbb{P}(H\text{ is not rainbow})\leq 1-\mathbb{P}(e\text{ and }e'\text{ are not rainbow})\,\color{blue}{<}\,1\,.$$ If $e(H)=0$, then the claim is vacuously true. If $e(H)=1$, then $k\geq 5$, which means $$4^{k-1}>3^k\,,$$ so $e(H)=1<\dfrac{4^{k-1}}{3^k}$, and (*) is true.