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

Question

Aicharya · Accepted Answer

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

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show…

Solution