Six dice are thrown 729 times. The number of times you expect atleast 3 dice to show either 5 or 6 is

Let P=Probability (showing 5 or 6)$\frac{2}{6}=\frac{1}{3}$

Now $\mathrm{q}=1-\mathrm{p}=\frac{2}{3}$

and n=6

Now P(at least 3 dice will show 5 or 6)

$$\begin{gathered} =1-\left[\mathrm{P}(X=0)+\mathrm{P}(X=1)+\mathrm{P}(X=2\right] \\ =1-\left[{}^6{\mathrm{c}}_0{}^{ } {\left(\frac{2}{3}\right)}^6 {\left(\frac{1}{3}\right)}^0+{}^6{\mathrm{c}}_1 {\left(\frac{1}{3}\right)}^1 {\left(\frac{2}{3}\right)}^5+{}^6{\mathrm{c}}_2 {\left(\frac{2}{3}\right)}^4 {\left(\frac{1}{3}\right)}^2\right] \\ =1-\left[1.{\left(\frac{2}{3}\right)}^6+6\left(\frac{1}{6}\right) {\left(\frac{2}{3}\right)}^5+15{\left(\frac{2}{3}\right)}^4 {\left(\frac{1}{3}\right)}^2\right] \\ =1-{\left(\frac{2}{3}\right)}^4 \left[\frac{4}{9}+\frac{12}{9}+\frac{15}{9}\right] \\ =1-\frac{16}{81}\times \frac{31}{9} \\ =1-\frac{496}{729} \\ =\frac{233}{729} \\ \therefore In 729 times,the expectation=729\left(\frac{233}{729}\right) \\ =233 \end{gathered}$$