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Question Number 131975 by liberty last updated on 10/Feb/21

$$\mathrm{Ninety}\mathrm{students},\mathrm{including}\mathrm{Joe}$$$$\mathrm{and}\mathrm{Jane}\mathrm{are}\mathrm{to}\mathrm{be}\mathrm{split}\mathrm{into}$$$$\mathrm{three}\mathrm{classes}\mathrm{of}\mathrm{equal}\mathrm{size}\mathrm{and}\mathrm{is}$$$$\mathrm{to}\mathrm{be}\mathrm{done}\mathrm{at}\mathrm{random}.\mathrm{What}\mathrm{is}$$$$\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{Joe}\mathrm{and}\mathrm{Jane}$$$$\mathrm{end}\mathrm{up}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{class}?$$

Commented by liberty last updated on 10/Feb/21

$$\mathrm{i}\mathrm{got}\mathrm{n}\left(\mathrm{S}\right)=\frac{90!}{30!.30!.30!}$$$$\mathrm{n}\left(\mathrm{A}\right)=3\times {\mathrm{C}}_{28}^{88}\times {\mathrm{C}}_{30}^{60}\times {\mathrm{C}}_{30}^{30}=3\times \frac{88!}{28!.60!}\times \frac{60!}{30!.30!}$$$$\mathrm{P}\left(\mathrm{A}\right)=\frac{3\times \frac{88!}{28!.}\times 30!}{90!}=\frac{3\times 30\times 29}{90\times 89}$$$$=\frac{29}{89}$$$$$$

Commented by liberty last updated on 10/Feb/21

$$\mathrm{hahaha}...{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{misread}\mathrm{the}\mathrm{question}$$

Answered by TheSupreme last updated on 10/Feb/21

$$Joeisintheclassx$$$$P(Jane\in x)=\frac{30-1}{90-1}=\frac{29}{89}$$$$$$

Commented by liberty last updated on 10/Feb/21

$$\mathrm{haha}\mathrm{its}\mathrm{general}\mathrm{formula}$$

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