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Category: Probability and Statistics

Question-58051

Question Number 58051 by Aditya789 last updated on 17/Apr/19 Answered by tanmay last updated on 17/Apr/19 $${sample}\:{space}\left\{{HHH},{HHT},{HTH},{HTT},{THH}\right. \\ $$$$\left.\:\:\:\:\:{THT},{TTH},{TTT}\right\} \\ $$$${favourabe}\:{result}\:{is}\left\{{HT}\bar {{H}},{THT}\right\} \\ $$$${so}\:{probablity}=\frac{\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\…

Question-58050

Question Number 58050 by Aditya789 last updated on 17/Apr/19 Answered by tanmay last updated on 17/Apr/19 $${D}_{\mathrm{1}} =\left\{\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:\right\} \\ $$$$\:\:{D}_{\mathrm{2}} =\left\{\mathrm{1}\:\mathrm{2}\:\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\right\} \\ $$$$\:\:{D}_{\mathrm{3}} \:=\left\{\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\right\} \\…

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-

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…

Question-57864

Question Number 57864 by mr W last updated on 13/Apr/19 Answered by mr W last updated on 14/Apr/19 $${to}\:{arrange}\:{n}\:{numbers}\:{there}\:{are}\:{n}!\:{ways}. \\ $$$${such}\:{that}\:{k}\:{numbers}\:{are}\:{together}\:{there} \\ $$$${are}\:\left({n}−{k}+\mathrm{1}\right)!{k}!\:{ways}. \\ $$$$\Rightarrow{p}=\frac{\left({n}−{k}+\mathrm{1}\right)!{k}!}{{n}!}=\frac{{n}−{k}+\mathrm{1}}{\frac{{n}!}{\left({n}−{k}\right)!{k}!}}=\frac{{n}−{k}+\mathrm{1}}{\:^{{n}}…

Question-57865

Question Number 57865 by mr W last updated on 13/Apr/19 Answered by MJS last updated on 13/Apr/19 $${n}=\mathrm{2}\:{p}=\mathrm{0} \\ $$$${n}=\mathrm{3}\:{p}=\mathrm{1}/\mathrm{3} \\ $$$${n}=\mathrm{4}\:{p}=\mathrm{3}/\mathrm{6}\:\left(=\mathrm{1}/\mathrm{2}\right) \\ $$$${n}=\mathrm{5}\:{p}=\mathrm{6}/\mathrm{10}\:\left(=\mathrm{3}/\mathrm{5}\right) \\…

let-P-k-C-k-4-k-1-determine-C-so-that-the-family-p-k-k-N-define-a-measure-of-probability-2-we-draw-a-number-in-N-according-to-the-probability-P-determined-the-probability-that-the

Question Number 123100 by pticantor last updated on 23/Nov/20 $$\boldsymbol{{let}}\:\boldsymbol{{P}}_{\boldsymbol{{k}}} =\frac{\boldsymbol{{C}}}{\boldsymbol{{k}}×\mathrm{4}^{\boldsymbol{{k}}\:} } \\ $$$$\left.\mathrm{1}\right)\:\boldsymbol{{determine}}\:\boldsymbol{{C}}\:\boldsymbol{{so}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{family}} \\ $$$$\left(\boldsymbol{{p}}_{\boldsymbol{{k}}} ,\boldsymbol{{k}}\in\mathbb{N}^{\ast} \right)\:\boldsymbol{{define}}\:\boldsymbol{{a}}\:\boldsymbol{{measure}}\:\boldsymbol{{of}}\:\boldsymbol{{probability}} \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{we}}\:\boldsymbol{{draw}}\:\boldsymbol{{a}}\:\boldsymbol{{number}}\:\boldsymbol{{in}}\:\mathbb{N}^{\ast} \:\boldsymbol{{according}}\:\boldsymbol{{to}}\:\boldsymbol{{the}}\: \\ $$$$\boldsymbol{{probability}}\:\boldsymbol{{P}}\:\boldsymbol{{determined}}\:\boldsymbol{{the}}\:\boldsymbol{{probability}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{number}} \\ $$$$\boldsymbol{{drawn}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{multiple}}\:\boldsymbol{{of}}\:\mathrm{4}…