Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$, then n is equal to _____ .

E₁ = denotes selection for 1^st bag
E₂ = denotes selection for 2^nd bag
$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{1}{2},\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{1}{2}$
A = selected balls are 1 red & 1 black
$\begin{array}{l}\mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)=\frac{{ }^3{\mathrm{C}}_1{\times }^1{\mathrm{C}}_1}{{ }^6{\mathrm{C}}_2}=\frac{1}{5}\\ \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)=\frac{{ }^3{\mathrm{C}}_1{\times }^2{\mathrm{C}}_1}{(\mathrm{n}+5{)}_{{\mathrm{C}}_2}}=\frac{12}{(\mathrm{n}+5)(\mathrm{n}+4)}\end{array}$
$\begin{array}{l}\mathrm{P}\left(\frac{{\mathrm{E}}_1}{\mathrm{A}}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_1\right)\times \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)}{\mathrm{P}\left({\mathrm{E}}_1\right)\times \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)+\mathrm{P}\left({\mathrm{E}}_2\right)\times \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_2}\right)}=\frac{6}{11}\\ \frac{6}{11}=\frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(\mathrm{n}+5)(\mathrm{n}+4)}}\\ \frac{\left(n+5\right)\left(n+4\right)}{\left(n+5\right)\left(n+4\right)+60}=\frac{6}{11}\Rightarrow 11\left(n^2+9n+20\right)=6\left(n^2+9n+80\right)\\ \Rightarrow n^2+9n-52=0\Rightarrow \left(n+13\right)\left(n-4\right)=0\\ \Rightarrow \mathrm{n}=4\end{array}$