Question Number 57952 by necx1 last updated on 14/Apr/19
$${An}\:{irregular}\:\mathrm{6}\:{faced}\:{die}\:{is}\:{thrown}\:{and} \\ $$$${the}\:{expectation}\:{that}\:{in}\:\mathrm{10}\:{throws}\:{it}\:{will} \\ $$$${give}\:{five}\:{even}\:{numbers}\:{is}\:{twice}\:{the} \\ $$$${expectation}\:{that}\:{it}\:{will}\:{give}\:{four}\:{even} \\ $$$${numbers}.{How}\:{many}\:{times}\:{in}\:\mathrm{15000} \\ $$$${sets}\:{of}\:\mathrm{10}\:{throws}\:{would}\:{you}\:{expect}\:{it} \\ $$$${to}\:{give}\:{one}\:{even}\:{number}? \\ $$
Answered by MJS last updated on 15/Apr/19
$$\frac{{p}^{\mathrm{5}} \left(\mathrm{1}−{p}\right)^{\mathrm{5}} }{{p}^{\mathrm{4}} \left(\mathrm{1}−{p}\right)^{\mathrm{6}} }=\mathrm{2} \\ $$$$\frac{{p}}{\mathrm{1}−{p}}=\mathrm{2}\:\Rightarrow\:{p}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$${p}\left(\mathrm{1}−{p}\right)^{\mathrm{9}} \begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{9}}\end{pmatrix}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{9}} }×\mathrm{10}=\frac{\mathrm{20}}{\mathrm{59049}} \\ $$$$\mathrm{15000}×\frac{\mathrm{20}}{\mathrm{59049}}=\frac{\mathrm{100000}}{\mathrm{19683}}\approx\mathrm{5} \\ $$