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

Probability to show 3 or 5 is
\(p=\frac{2}{6}=\frac{1}{3}\)
\(\therefore q=1-\frac{1}{3}=\frac{2}{3} \)
\(P(X \geq 2)\)
\(={ }^4 \mathrm{C}_2\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2+{ }^4 \mathrm{C}_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^1+{ }^4 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^0 \)
\( =6\left(\frac{1}{9}\right)\left(\frac{4}{9}\right)+4\left(\frac{1}{27}\right)\left(\frac{2}{3}\right)+1\left(\frac{1}{81}\right) \)
\( =\frac{24+8+1}{81} \)
\( =\frac{33}{81}=\frac{11}{27}\)
Four fair dice are thrown independently 27 times.
\(\therefore \quad\) Expected number \(=27 \times \frac{11}{27}=11\)