Question Number 12306 by Gaurav3651 last updated on 18/Apr/17
$${Two}\:{similar}\:{boxes}\:{B}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right){contain} \\ $$$$\left({i}+\mathrm{1}\right){red}\:{and}\:\left(\mathrm{5}−{i}−\mathrm{1}\right)\:{black}\:{balls}. \\ $$$${One}\:{box}\:{is}\:{chosen}\:{at}\:{random}\:{and} \\ $$$${two}\:{balls}\:{are}\:{drawn}\:{randomly}. \\ $$$${what}\:{is}\:{the}\:{probability}\:{that}\:{both} \\ $$$${balls}\:{are}\:{of}\:{different}\:{colours}? \\ $$$$\left({a}\right)\:\:\mathrm{1}/\mathrm{2} \\ $$$$\left({b}\right)\:\:\mathrm{3}/\mathrm{10} \\ $$$$\left({c}\right)\:\:\mathrm{2}/\mathrm{5} \\ $$$$\left({d}\right)\:\:\mathrm{3}/\mathrm{5} \\ $$
Answered by mrW1 last updated on 19/Apr/17
$$\mathrm{1}×\frac{\mathrm{2}×\mathrm{3}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow\left({d}\right) \\ $$
Commented by Gaurav3651 last updated on 19/Apr/17
$${Sir}\:{can}\:{you}\:{please}\:{explain}\:{the} \\ $$$${solution}. \\ $$
Commented by mrW1 last updated on 19/Apr/17
$${The}\:{answer}\:{should}\:{be}\:\mathrm{1}×\:\frac{\mathrm{6}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}},\:{since} \\ $$$${it}\:{makes}\:{no}\:{difference}\:{which}\:{box}\:{is} \\ $$$${choosen}. \\ $$$$\mathrm{5}\:{balls}\:{in}\:{each}\:{box},\:\mathrm{3}\:{balls}\:{in}\:{colour}\:{A} \\ $$$${and}\:\mathrm{2}\:{balls}\:{in}\:{colour}\:{B}. \\ $$$$ \\ $$$${possibilities}\:{to}\:{take}\:\mathrm{2}\:{from}\:\mathrm{5}\:{balls}:\: \\ $$$${C}\left(\mathrm{5},\mathrm{2}\right)=\mathrm{10} \\ $$$${possibilities}\:{to}\:{take}\:\mathrm{2}\:{balls}\:{in}\:{different}\:{colours}:\: \\ $$$$\mathrm{2}×\mathrm{3}=\mathrm{6} \\ $$$$ \\ $$$$\Rightarrow\mathrm{1}×\frac{\mathrm{6}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Commented by Gaurav3651 last updated on 19/Apr/17
$${Thanks}\:{a}\:{lot}\:{sir} \\ $$$$ \\ $$