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if-three-numbers-are-drawn-at-random-successively-without-replacement-from-a-set-S-1-2-10-then-probability-that-the-minimum-of-the-choosen-number-is-3-or-their-maximum-is-7-Answer-11

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Question Number 40550 by vajpaithegrate@gmail.com last updated on 24/Jul/18
if three numbers are drawn at random successively without replacement from a set S={1,2,......10}then probability that the minimum of the choosen number is 3 or their maximum is 7 Answer=((11)/(40))
$$\mathrm{if}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{successively}\:\mathrm{without}\:\mathrm{replacement}\:\mathrm{from} \\ $$$$\mathrm{a}\:\mathrm{set}\:\mathrm{S}=\left\{\mathrm{1},\mathrm{2},……\mathrm{10}\right\}\mathrm{then}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{choosen}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{3}\:\mathrm{or}\:\mathrm{their}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{7}\:\:\:\:\:\:\:\: \\ $$$$\:\mathrm{Answer}=\frac{\mathrm{11}}{\mathrm{40}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jul/18
P(A)=probablity to choose three number where 3 is minimum so choosing 3 and any two number from(4,5, 6,7,8,9,10) P(A)((1c_1 ×7c_2 )/(10c_3 ))=((1×((7×6)/2))/((10×9×8)/(3×2)))=((21)/(120))=(7/(40)) P(B)=probabligy to choose three number where 7 is maximum so choosing 7 and any two number from (1,2,3,4,5,6) P(B)((1c_1 ×6c_2 )/(10c_3 ))=((15)/((10×9×8)/(3×2)))=((15)/(120))=(5/(40)) P(A)orP(B) P(A)∪P(B)=P(A)+P(B)−P(A)∩P(B) calculaion of P(A)∩P(B) (3,4,5,6,7) choosing 3 and 7 and any one from (4,5,6) P(A)∩P(B)=((1c_1 ×1c_1 ×3c_1 )/(10c_3 ))=(3/(120))=(1/(40)) so required results=(7/(40))+(5/(40))−(1/(40))=((11)/(40))
$${P}\left({A}\right)={probablity}\:{to}\:{choose}\:{three}\:{number} \\ $$$${where}\:\mathrm{3}\:{is}\:{minimum} \\ $$$${so}\:{choosing}\:\mathrm{3}\:{and}\:{any}\:{two}\:{number}\:{from}\left(\mathrm{4},\mathrm{5},\right. \\ $$$$\left.\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right) \\ $$$${P}\left({A}\right)\frac{\mathrm{1}{c}_{\mathrm{1}} ×\mathrm{7}{c}_{\mathrm{2}} }{\mathrm{10}{c}_{\mathrm{3}} }=\frac{\mathrm{1}×\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}}}{\frac{\mathrm{10}×\mathrm{9}×\mathrm{8}}{\mathrm{3}×\mathrm{2}}}=\frac{\mathrm{21}}{\mathrm{120}}=\frac{\mathrm{7}}{\mathrm{40}} \\ $$$${P}\left({B}\right)={probabligy}\:{to}\:{choose}\:{three}\:{number} \\ $$$${where}\:\mathrm{7}\:{is}\:{maximum} \\ $$$${so}\:{choosing}\:\mathrm{7}\:{and}\:{any}\:{two}\:{number}\:{from} \\ $$$$\left(\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right) \\ $$$${P}\left({B}\right)\frac{\mathrm{1}{c}_{\mathrm{1}} ×\mathrm{6}{c}_{\mathrm{2}} }{\mathrm{10}{c}_{\mathrm{3}} }=\frac{\mathrm{15}}{\frac{\mathrm{10}×\mathrm{9}×\mathrm{8}}{\mathrm{3}×\mathrm{2}}}=\frac{\mathrm{15}}{\mathrm{120}}=\frac{\mathrm{5}}{\mathrm{40}} \\ $$$${P}\left({A}\right){orP}\left({B}\right) \\ $$$${P}\left({A}\right)\cup{P}\left({B}\right)={P}\left({A}\right)+{P}\left({B}\right)−{P}\left({A}\right)\cap{P}\left({B}\right) \\ $$$${calculaion}\:{of}\:{P}\left({A}\right)\cap{P}\left({B}\right) \\ $$$$\left(\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7}\right)\:\:\:{choosing}\:\mathrm{3}\:{and}\:\mathrm{7}\:{and}\:{any}\:{one}\:{from} \\ $$$$\left(\mathrm{4},\mathrm{5},\mathrm{6}\right) \\ $$$${P}\left({A}\right)\cap{P}\left({B}\right)=\frac{\mathrm{1}{c}_{\mathrm{1}} ×\mathrm{1}{c}_{\mathrm{1}} ×\mathrm{3}{c}_{\mathrm{1}} }{\mathrm{10}{c}_{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{120}}=\frac{\mathrm{1}}{\mathrm{40}} \\ $$$${so}\:{required}\:{results}=\frac{\mathrm{7}}{\mathrm{40}}+\frac{\mathrm{5}}{\mathrm{40}}−\frac{\mathrm{1}}{\mathrm{40}}=\frac{\mathrm{11}}{\mathrm{40}} \\ $$
Commented by vajpaithegrate@gmail.com last updated on 25/Jul/18
tnq sir
$$\mathrm{tnq}\:\mathrm{sir} \\ $$

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