For this problem, use the Negative Binomial Distribution when looking for the number of trials till the 2nd success. p is determined in the first answer.
P(7 or 11 on one roll) = 8/36 = 2/9, using equally likely outcomes.
\(\displaystyle \mu = \frac{r}{p} = \frac{2}{2/9} = 9\)
\(\displaystyle P( X = 12) = \binom{11}{1} (7/9)^10 \cdot (2/9)^2 = 0.044\)
\begin{align*}
P(X \ge 4) & = 1- P(x \le 3) \\
& = 1 - [ f(2) + f(3) ]\\
& = 1 - \left[ \binom{1}{1} (2/9)^2 + \binom{2}{1} (7/9)^1 \cdot (2/9)^2 \right ]\\
& = 1 - 0.1262 = 0.8738.
\end{align*}