**Question:**

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

**Solution:**

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

$\mathrm{p}=\frac{2}{6}=\frac{1}{3}$

$\therefore \quad \mathrm{q}=1-\frac{1}{3}=\frac{2}{3}$ (not showing 3 or 5 )

Experiment is performed with 4 dices independently.

$\therefore$ Their binomial distribution is

$(q+p)^{4}=(q)^{4}+{ }^{4} C_{1} q^{3} p+{ }^{4} C_{2} q^{2} p^{2}+{ }^{4} C_{3} q p^{3}$

$+{ }^{4} C_{4} p^{4}$

$\therefore$ In one throw of each dice probability of showing 3 or 5 at least twice is

$=\mathrm{p}^{4}+{ }^{4} \mathrm{C}_{3} \mathrm{qp}^{3}+{ }^{4} \mathrm{C}_{2} \mathrm{q}^{2} \mathrm{p}^{2}$

$=\frac{33}{81}$

$\therefore$ Such experiment performed 27 times

$\therefore$ so expected out comes $=\mathrm{np}$

$=\frac{33}{81} \times 27$

$=11$