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

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