Question Number 87253 by peter frank last updated on 03/Apr/20
$${Three}\:{pair}\:{of}\:{socks}\:{are} \\ $$$${placed}\:{in}\:{a}\:{box}.{If}\:{two} \\ $$$${socks}\:{are}\:{drawn}\:{at} \\ $$$${random}\:{from}\:{the}\:{box} \\ $$$${What}\:{is}\:{the}\:{probability} \\ $$$$\left({a}\right){of}\:{drawing}\:\:{a}\:{match} \\ $$$${pair} \\ $$$$\left({b}\right){of}\:{drawing}\:{a}\:{socks} \\ $$$${for}\:{the}\:{left}\:{and}\:{right} \\ $$$${feet} \\ $$$$\left({c}\right){of}\:{drawing}\:{two}\:{socks}\:{of} \\ $$$${the}\:{right}\:{feet} \\ $$$$\left.{d}\right){drawing}\:{two}\:{socks}\:{of} \\ $$$${left}\:{feet} \\ $$$$\left({e}\right){drawing}\:{socks}\:{of}\:{the} \\ $$$${same}\:{feet} \\ $$$$ \\ $$
Answered by malwaan last updated on 03/Apr/20
$$\left({a}\right)\:\frac{\mathrm{1}+\mathrm{1}+\mathrm{1}}{{C}_{\mathrm{2}} ^{\mathrm{6}} }\:=\frac{\mathrm{3}}{\mathrm{15}}\:=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\left({b}\right)\:\frac{{C}_{\mathrm{1}} ^{\mathrm{3}} ×{C}_{\mathrm{1}} ^{\mathrm{3}} }{{C}_{\mathrm{2}} ^{\mathrm{6}} }\:=\frac{\mathrm{3}×\mathrm{3}}{\mathrm{15}}=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\left({c}\right)\:\frac{{C}_{\mathrm{2}} ^{\mathrm{3}} \:}{{C}_{\mathrm{2}} ^{\mathrm{6}} }\:=\frac{\mathrm{3}}{\mathrm{15}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\left({d}\right)\:=\frac{\mathrm{1}}{\mathrm{5}}\:\left(==\left({c}\right)\Lsh\right) \\ $$$$\left({e}\right)\:\frac{{C}_{\mathrm{2}} ^{\mathrm{3}} \:+{C}_{\mathrm{2}} ^{\mathrm{3}} }{{C}_{\mathrm{2}} ^{\mathrm{6}} }\:=\frac{\mathrm{3}+\mathrm{3}}{\mathrm{15}}=\frac{\mathrm{6}}{\mathrm{15}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$