# 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 _______. Option: 1 9 Option: 2 10 Option: 3 11 Option: 4 12

4 dice are independently thrown. Each die has the probability to show 3 or 5 is

\begin{aligned} &{p}=\frac{2}{6}=\frac{1}{3}\\ &\therefore \quad q=1-\frac{1}{3}=\frac{2}{3}(\text { not showing } 3 \text { or } 5) \end{aligned}

The experiment is performed with 4 dices independently

Therefore, Their binomial distribution is

\begin{aligned} &(q+p)^{4}=(q)^{4}+{ }^{4} C_{1} q^{3} p+{ }^{4} C_{2} q^{2} p^{2}+{ }^{4} C_{3} q p^{3}+{ }^{4} C_{4} P^{4}\\ &\therefore \text { In one throw of each dice probability of showing } 3 \text { or } 5 \text { at least twice is }\\ &=\mathrm{p}^{4}+{ }^{4} \mathrm{C}_{3} \mathrm{qp}^{3}+{ }^{4} \mathrm{C}_{2} \mathrm{q}^{2} \mathrm{p}^{2}\\ &=\frac{33}{81} \end{aligned}

$\\\therefore\text{ Such experiment performed 27 times} \\\therefore\text{so expected out comes }=\mathrm{np} \\=\frac{33}{81} \times 27=11$

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