$\mathbf{\underline{Answer:}\Rightarrow} \mathbf{B}$
$\mathbf{\underline{Explanation:}\Rightarrow} \mathbf{}$
Total ways in which $\color {green} {\Large {2\mathrm n}}$ balls can be chosen where $\color {red} {\mathrm {\Large n}}$ are of $ \color {red} {\text{Red Colour}}$ and other $\color {blue}{\Large {\mathrm n}}$ are of $\color {blue} {\text{Blue Colour}}$ = $\dfrac{(2\mathrm n)!}{\mathrm n!.\mathrm n!}$ = $\dfrac{2}{\mathrm n!}$ = $\Large \binom{2\mathrm n}{\mathrm {C_n}}$
For the person to remain dry he should have chosen more Red balls than the blue balls which is given by $\Large \color {darkgreen} {\mathrm {C_n}}$.
$\therefore$ The answer $=\color {green} {\Large {\dfrac{\mathrm {C_n}}{\binom{2\mathrm n}{\mathrm {C_n}}}}}$
$\therefore $ B is the correct option.