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Question Number 99225 by BigBoss last updated on 19/Jun/20

$$\mathrm{A}\mathrm{bag}\mathrm{contains}4\mathrm{tickets}\mathrm{numbered}1,2,$$$$3,4\mathrm{and}\mathrm{another}\mathrm{bag}\mathrm{contains}6\mathrm{tickets}$$$$\mathrm{numbered}2,4,6,7,8,9.\mathrm{One}\mathrm{bag}\mathrm{is}$$$$\mathrm{chosen}\mathrm{and}\mathrm{a}\mathrm{ticket}\mathrm{is}\mathrm{drawn}.\mathrm{The}$$$$\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{ticket}\mathrm{bears}\mathrm{the}$$$$\mathrm{number}4\mathrm{is}$$

Answered by john santu last updated on 19/Jun/20

$$\mathrm{p}\left(\mathrm{A}\right)=\frac{1}{2}\times \{\frac{1}{4}+\frac{1}{6}\}=$$$$=\frac{1}{2}\times \frac{5}{12}=\frac{5}{24}◼$$

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