Let $P=$ probability (showing 5 or 6$)=2 / 6=1 / 3$ $q=1-p=1-1 / 3=2 / 3$ $n=6$ and $r=3$

Also $p(x=r)=$ probability (at least 3 dice will show 5 or 6 in one trial) Using the 'complement ' theorem $$ \begin{array}{l} p(x=r)=1-[p(x=0)+p(x=1)+p(x=2)] \\ =1-\left[{ }^{n} C_{0}\left(\frac{2}{3}\right)^{6}\left(\frac{1}{3}\right)^{0}+{ }^{6} C_{1}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)^{1}+{ }^{6} C_{2}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{2}\right] \\ =1-\left[1 \cdot\left(\frac{2}{3}\right)^{6}\left(\frac{1}{3}\right)^{0}+6\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)^{1}+15\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{2}\right] \\ =1-\left(\frac{2}{3}\right)^{4}\left[\frac{4}{9}+\frac{12}{9}+\frac{15}{9}\right] \\ =1-\frac{16}{9} \times \frac{31}{9} \\ =1-\frac{496}{729} \\ =\frac{233}{729} \end{array} $$ Therefore in 729 trials, the expression $=729 \times \frac{233}{729}=233$