SOLUTION
From the question, we want to find the probability of selecting a slip of paper with the letter t on it, then selecting the letter r, with replacement.
This means the probability of selecting t and r.
That is
[tex]P(t)\times P(r)[/tex]There are 12 letters (total outcomes). And there are 2r and 2t.
So probability becomes
[tex]\begin{gathered} \frac{possible\text{ outcome}}{\text{total outcome }} \\ P(t)=\frac{2}{12} \\ P(r)=\frac{2}{12} \\ P(t)\times P(r)=\frac{2}{12}\times\frac{2}{12} \\ =\frac{1}{6}\times\frac{1}{6} \\ =\frac{1}{36} \end{gathered}[/tex]Hence the answer is
[tex]\frac{1}{36}[/tex]