I can find the expected value of the number of results that appear atleast once. I need to either find the expected value of that number squared or the probability of having $k$ results appear atleast once. and then use $Var=\mathbb{E}[X^2]-\mathbb{E}^2[X]$
I would really appreciate some help!
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For $k=1,2,3,4,5,6$ let $B_{k}$ take value $1$ if face $k$ appears at least once and let it take value $0$ otherwise.
Now observe that $B:=\sum_{k=1}^{6}B_{k}$ equals the number of faces that appear at least once.
Here:$$\mathsf{Var}\left(B\right)=\mathsf{Cov}\left(B,B\right)=\sum_{i=1}^{6}\sum_{j=1}^{6}\mathsf{Cov}\left(B_{i},B_{j}\right)=6\mathsf{Cov}\left(B_{1},B_{1}\right)+30\mathsf{Cov}\left(B_{1},B_{2}\right)=$$$$6\mathsf{Var}B_{1}+30\mathsf{Cov}\left(B_{1},B_{2}\right)$$
Can you find $\mathsf{Var}B_1$ and $\mathsf{Cov}(B_1,B_2)$ yourself?
Hint: Let $X$ be the number of numbers that appear at least once. Obviously, $X$ takes on values in $\{1,2,3,4,5,6\}$. Calculate, if you have not already done so, $\mathbb{P}(X=i)$ for all $i=1,2,\ldots,6$. Calculate $$\mathbb{E}X^2=\sum_{i=1}^6 i^2 \mathbb{P}(X=i).$$ Since you already know $\mathbb{E}X$, then you can calculate the variance as follows: $$\operatorname{Var}X=\mathbb{E}X^2-(\mathbb{E}X)^2.$$