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Category: Probability and Statistics

A-dice-is-cast-twice-and-the-sum-of-the-appearing-numbers-is-10-The-probability-that-the-number-5-has-appeared-at-least-once-is-

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Question Number 201969 by BaliramKumar last updated on 17/Dec/23 $$\mathrm{A}\:\mathrm{dice}\:\mathrm{is}\:\mathrm{cast}\:\mathrm{twice},\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{appearing}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{10}. \\ $$$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\mathrm{5}\:\mathrm{has}\: \\ $$$$\mathrm{appeared}\:\mathrm{at}\:\mathrm{least}\:\mathrm{once}\:\mathrm{is}. \\ $$ Answered by Frix last updated on 17/Dec/23…

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Question-201527

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Question Number 201527 by cortano12 last updated on 08/Dec/23 Answered by AST last updated on 08/Dec/23 $$#\left(\mathrm{6}\:{or}\:\mathrm{8}\right)=#\left(\mathrm{6}\right)+#\left(\mathrm{8}\right)−#\left(\mathrm{6}{n}\mathrm{8}\right) \\ $$$$#\left(\mathrm{6}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{6}}\rfloor=\mathrm{333};#\left(\mathrm{8}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{8}}\rfloor=\mathrm{250} \\ $$$$#\left(\mathrm{6}{n}\mathrm{8}\right)=#\left(\mathrm{24}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{24}}\rfloor=\mathrm{83} \\ $$$$\Rightarrow#\left(\mathrm{6}\:{or}\:\mathrm{8}\right)=\mathrm{500} \\ $$$$\Rightarrow{Probability}=\frac{\mathrm{2000}−\mathrm{500}}{\mathrm{2000}}=\frac{\mathrm{3}}{\mathrm{4}}…

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