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Question Number 131631 by liberty last updated on 07/Feb/21

$$\mathrm{Selecting}\mathrm{randomly}\mathrm{an}\mathrm{integer}$$$$\mathrm{from}1\mathrm{to}2000,\mathrm{find}\mathrm{the}\mathrm{probability}$$$$\mathrm{that}\mathrm{neither}6\mathrm{nor}8\mathrm{divides}\mathrm{the}$$$$\mathrm{integer}.$$

Answered by som(math1967) last updated on 07/Feb/21

$$probabilitydivisibleby6$$$$\frac{333}{2000}$$$$[Inthesereies6,12....1998$$$$1998=6+(n-1)\times 6\therefore n=333]$$$$probabilitydivisibleby8$$$$=\frac{250}{2000}[8+(n-1)\times 8=2000$$$$\therefore n=250]$$$$probabilitydivisiblebyboth$$$$6,8=\frac{83}{2000}$$$$[1992=24+(n-1)\times 24$$$$n=83]$$$$\therefore probabilitydivisibleby$$$$6or8=\frac{333}{2000}+\frac{250}{2000}-\frac{83}{2000}$$$$=\frac{500}{2000}$$$$\therefore neither6nor8$$$$1-\frac{500}{2000}=\frac{1500}{2000}=\frac{3}{4}$$

Commented by liberty last updated on 07/Feb/21

$$\mathrm{typo}\mathrm{sir}\mathrm{it}\mathrm{should}\mathrm{be}\frac{250}{2000}\left[8\mathrm{n}\right]=2000$$$$\mathrm{n}=250$$

Commented by som(math1967) last updated on 07/Feb/21

$$yessir,thankyou$$

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