We consider three fair dice which are thrown simultanously. In other words, one repetition of the experiment means that we have thrown three dice.
Use Chebyshev inequality $P(|X-\mathbb{E}(X)|\geq \epsilon)\leq \frac{\mathbb{V}(X)}{\epsilon^2}$, where $\epsilon>0$, to find the minimum number of repetitions of the experiment such that the probability of getting at least two $3$'s is $0.25$ or more. We can assume that each repitition is independent from others.
I know that this question seems odd because it is easy to do it without Chebyshev inequality. However, our professor wants us to apply Chebyshev inequality.
My approach:
Let be $X_i$ a random variable with $1\leq i\leq n$, such that $X_i=1$, if at least two $3$'s show up and otherwise $X_i=0$. It is easy to see that $P(X_i=1)=\frac{2}{27}$ and $P(X_i=0)=\frac{25}{27}$. Then, we know $Y:=\sum\limits_{i=1}^{n}X_i$ obeys a Binomial distribution and $\mathbb{E}(Y)=\frac{2n}{27}$ and $\mathbb{V}(Y)=\frac{50n}{27^2}$. To answer the question we must find a $n$ such that $P(Y\geq 1)$.
In order to apply Chebyshev inequality we have to do some tricky manipulations and assume that $n$ is large enough such that $\mathbb{E}(Y)-1=\frac{2n}{27}-1>0$.
\begin{align*} &1.) ~P(Y\geq 1)\geq 0.25\iff P(Y<1)\leq 0.75\\ &~\\ &2.)~P(Y<1)=P(Y-\mathbb{E}(Y)\leq 1-\mathbb{E}(Y))=P(\mathbb{E}(Y)-Y\geq \mathbb{E}(Y)-1)\\ &\leq P(|\mathbb{E}(Y)-Y|\geq \mathbb{E}(Y)-1)\leq \frac{\mathbb{V}(Y)}{(\mathbb{E}(Y)-1)^2}\\ &\implies P(Y<1)\leq \frac{\mathbb{V}(Y)}{(\mathbb{E}(Y)-1)^2}=\frac{\frac{50n}{27^2}}{\left(\frac{2n}{27}-1\right)^2}. \end{align*}
In equation 2.) I have used the subset property and applied Chebyshev inequality. Long story short, we are looking for a $n\in\mathbb{N}$ such that $\frac{\frac{50n}{27^2}}{\left(\frac{2n}{27}-1\right)^2}\leq 0.75$. After some manipulations of the inequality we see that $n\geq 39$.
My tutor deducted almost $50$% of the points just mentioning that I can't do it this way and it is not clear if the $n$ I found is truely the smallest number of repitition if we use Chebyshev inequality. To be honest I don't know what she meant and I don't see any mistakes. Maybe someone else can explain to me what is wrong with my solution.