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Question Number 40550 by vajpaithegrate@gmail.com last updated on 24/Jul/18

$$\mathrm{if}\mathrm{three}\mathrm{numbers}\mathrm{are}\mathrm{drawn}\mathrm{at}\mathrm{random}$$$$\mathrm{successively}\mathrm{without}\mathrm{replacement}\mathrm{from}$$$$\mathrm{a}\mathrm{set}\mathrm{S}=\{1,2,......10\}\mathrm{then}\mathrm{probability}\mathrm{that}$$$$\mathrm{the}\mathrm{minimum}\mathrm{of}\mathrm{the}\mathrm{choosen}\mathrm{number}\mathrm{is}$$$$3\mathrm{or}\mathrm{their}\mathrm{maximum}\mathrm{is}7$$$$\mathrm{Answer}=\frac{11}{40}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jul/18

$$P\left(A\right)=probablitytochoosethreenumber$$$$where3isminimum$$$$sochoosing3andanytwonumberfrom(4,5,$$$$6,7,8,9,10)$$$$P\left(A\right)\frac{1c_1\times 7c_2}{10c_3}=\frac{1\times \frac{7\times 6}{2}}{\frac{10\times 9\times 8}{3\times 2}}=\frac{21}{120}=\frac{7}{40}$$$$P\left(B\right)=probabligytochoosethreenumber$$$$where7ismaximum$$$$sochoosing7andanytwonumberfrom$$$$(1,2,3,4,5,6)$$$$P\left(B\right)\frac{1c_1\times 6c_2}{10c_3}=\frac{15}{\frac{10\times 9\times 8}{3\times 2}}=\frac{15}{120}=\frac{5}{40}$$$$P\left(A\right)orP\left(B\right)$$$$P\left(A\right)\cup P\left(B\right)=P\left(A\right)+P\left(B\right)-P\left(A\right)\cap P\left(B\right)$$$$calculaionofP\left(A\right)\cap P\left(B\right)$$$$(3,4,5,6,7)choosing3and7andanyonefrom$$$$(4,5,6)$$$$P\left(A\right)\cap P\left(B\right)=\frac{1c_1\times 1c_1\times 3c_1}{10c_3}=\frac{3}{120}=\frac{1}{40}$$$$sorequiredresults=\frac{7}{40}+\frac{5}{40}-\frac{1}{40}=\frac{11}{40}$$

Commented by vajpaithegrate@gmail.com last updated on 25/Jul/18

$$\mathrm{tnq}\mathrm{sir}$$

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