Let be $X_i$ a Bernoulli distributed random variable, such that $P(X_i=1)=\frac{2}{27}$ and $P(X_i=0)=\frac{25}{27}$. Then, $X:=\sum\limits_{i=1}^nX_i$ obeys a binomial distribution, where $\mathbb{E}(X)=\frac{2n}{27}$ and $\mathbb{V}(X)=\frac{50n}{27^2}$.
Use Chebycheff inequality $P(|X-\mathbb{X}(X)|\geq \epsilon)\leq\frac{\mathbb{V}(X)}{\epsilon^2}$ to estimate the smallest $n$ such that the probability of at least one success is $\geq 0.25$.
(Do use Chebycheff inequality to find an estimate and don't calculate the true $n$.)
First approach:
We are trying to show that $P(X\geq 1)\geq 0.25$ which is equivalent to $P(X<1)<0.75$. We see that \begin{align*} &P(X< 1)=P(X-\mathbb{E}(X)<1-\mathbb{E}(X))=P(\mathbb{E}(X)-X>\mathbb{E}(X)-1)\leq P(|\mathbb{E}(X)-X|>\mathbb{E}(X)-1). \end{align*} If we assume $\mathbb{E}(X)-1=\frac{2n}{27}-1>0$, then Chebycheff inequality (CI) yields \begin{align*} &P(X< 1)\leq P(|\mathbb{E}(X)-X|>\mathbb{E}(X)-1)\overset{\text{CI}}{\leq}\frac{\mathbb{V}(X)}{(\mathbb{E}(X)-1)^2}\leq 0.75\\ &\implies \frac{\frac{50n}{27^2}}{(\frac{2n}{27}-1)^2}\leq 0.75\implies n\geq 39. \end{align*}
Second approach:
We start with \begin{align*} P(|X-\mathbb{E}(X)|\geq \epsilon)\leq\frac{\mathbb{V}(X)}{\epsilon^2}\iff P(|X-\mathbb{E}(X)|\leq \epsilon)\geq1-\frac{\mathbb{V}(X)}{\epsilon^2} \end{align*} and because of $\{|X-\mathbb{E}(X)|\leq \epsilon\}\subseteq \{\mathbb{E}(X)-X\leq \epsilon\}=\{X\geq \mathbb{E}(X)- \epsilon\}$, it follows
\begin{align*} P(X\geq \mathbb{E}(X)-\epsilon)\geq1-\frac{\mathbb{V}(X)}{\epsilon^2}. \end{align*} If we assume $\mathbb{E}(X)-\epsilon>0$ and look at $1-\frac{\mathbb{V}(X)}{\epsilon^2}\geq 0.25$, then tinkering around with $n$ and $\epsilon>0$ simultanously delivers $n=18$ (and $\epsilon\approx1.31$).
Why do the two approaches produce different results? As the first approach yields a much larger $n$ there must be an issue.