Question Number 113605 by bobhans last updated on 14/Sep/20
$$\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{6}\:\mathrm{red},\:\mathrm{5}\:\mathrm{white}\:\mathrm{and}\:\mathrm{4}\:\mathrm{black}\:\mathrm{balls} \\ $$$$\mathrm{If}\:\mathrm{twl}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{drawn}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{none}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{red}\:? \\ $$$$ \\ $$
Commented by bemath last updated on 14/Sep/20
$${method\_}\mathrm{2} \\ $$$${p}\left({A}\right)=\frac{\mathrm{9}}{\mathrm{15}}×\frac{\mathrm{8}}{\mathrm{14}}\:=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{4}}{\mathrm{7}}=\frac{\mathrm{12}}{\mathrm{35}} \\ $$
Answered by bemath last updated on 14/Sep/20
$${method\_}\mathrm{1} \\ $$$${n}\left({S}\right)=\:{C}_{\mathrm{2}} ^{\mathrm{15}} \:=\:\frac{\mathrm{15}×\mathrm{14}}{\mathrm{2}×\mathrm{1}}\:=\:\mathrm{105} \\ $$$${none}\:{of}\:{them}\:{are}\:{red}\rightarrow\left(\mathrm{2}{W}\right),\left(\mathrm{2}{B}\right),\left(\mathrm{1}{W},\mathrm{1}{B}\right) \\ $$$${n}\left({A}\right)={C}_{\mathrm{2}} ^{\mathrm{5}} +{C}_{\mathrm{2}} ^{\mathrm{4}} +{C}_{\mathrm{1}} ^{\mathrm{5}} ×{C}_{\mathrm{1}} ^{\mathrm{4}} \\ $$$${n}\left({A}\right)=\:\mathrm{10}+\mathrm{6}+\mathrm{20}=\mathrm{36} \\ $$$${so}\:{p}\left({A}\right)\:=\:\frac{\mathrm{36}}{\mathrm{105}}\:=\:\frac{\mathrm{12}}{\mathrm{35}} \\ $$