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
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 $=\mathrm{np}$
$=\frac{33}{81} \times 27$
$=11$
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