Mathematics - Probability Question with Solution | TestHub

Let$\mathrm{A} :$one ball is white and other is red
${\mathrm{E}}_1 :$both balls are from box${\mathrm{B}}_1$
${\mathrm{E}}_2 :$both ball are from box${\mathrm{B}}_2$
${\mathrm{E}}_3 :$both ball are from box${\mathrm{B}}_3$
Hence,$\text{P}\left(\text{required}\right)=\text{P}\left(\frac{{\text{E}}_2}{\text{A}}\right)$
$=\frac{\text{P}\left(\frac{\text{A}}{{\text{E}}_2}\right)·\text{P}\left({\text{E}}_2\right)}{\text{P}\left(\frac{\text{A}}{{\text{E}}_1}\right)·\text{P}\left({\text{E}}_1\right)+\text{P}\left(\frac{\text{A}}{{\text{E}}_2}\right)·\text{P}\left({\text{E}}_2\right)+\text{P}\left(\frac{\text{A}}{{\text{E}}_3}\right)·\text{P}\left({\text{E}}_3\right)}$
$=\frac{\frac{{}^2{\text{C}}_1\times {}^3{\text{C}}_1}{{}^9{\text{C}}_2}\times \frac{1}{3}}{\frac{{}^1{\text{C}}_1\times {}^3{\text{C}}_1}{{}^6{\text{C}}_2}\times \frac{1}{3}+\frac{{}^2{\text{C}}_1\times {}^3{\text{C}}_1}{{}^9{\text{C}}_2}\times \frac{1}{3}+\frac{{}^3{\text{C}}_1\times {}^4{\text{C}}_1}{{}^{12}{\text{C}}_2}\times \frac{1}{3}}$
$=\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}}$
$=\frac{55}{181}$