Question

Four fair dice are thrown independently $27$ times. Then the expected number of times, at least two dice show up a three or five, is

Answer 1

$\text{P}\left(\text{at 2, 3 or 4}\right)={}^4{C}_2{\left(\frac{2}{6}\right)}^2{\left(\frac{4}{6}\right)}^2+{}^4{C}_3{\left(\frac{2}{6}\right)}^3{\left(\frac{4}{6}\right)}^1+{}^4{C}_4{\left(\frac{2}{6}\right)}^4$
$=6\times \frac{1}{9}\times \frac{4}{9}+4\times \frac{1}{27}\times \frac{2}{3}+\frac{1}{81}$
$=\frac{33}{81}=\frac{11}{27}$
Such expression is performed $27$ times
So, expected outcome $=np$
$=\frac{11}{27}\times 27=11$