Bag A contains 2 white, 1 black and 3 red balls and bag B contains...

Let ${E}_{1}$ event denotes the selection for ${1}^{st}$ bag

and ${E}_{2}$ event denotes the selection for ${2}^{nd}$ bag

Given $P\left({E}_{1}\right) = \dfrac{1}{2}, P\left({E}_{2}\right) = \dfrac{1}{2}$

Let the event $A =$ selected balls are $1$ red & $1$ black

$P\left(\dfrac{A}{{E}_{1}}\right) = \dfrac{C1 3 \times C1 1}{C2 6} = \dfrac{1}{5}$

$P\left(\dfrac{A}{{E}_{2}}\right) = \dfrac{C1 3 \times C1 2}{{\left(n + 5\right)}_{{C}_{2}}} = \dfrac{12}{\left(n + 5\right)\left(n + 4\right)}$

$P\left(\dfrac{{E}_{1}}{A}\right) = \dfrac{P\left({E}_{1}\right) \times P\left(\dfrac{A}{{E}_{1}}\right)}{P\left({E}_{1}\right) \times P\left(\dfrac{A}{{E}_{1}}\right) + P\left({E}_{2}\right) \times P\left(\dfrac{A}{{E}_{2}}\right)}$

$⇒ \dfrac{\dfrac{1}{2} \times \dfrac{1}{5}}{\dfrac{1}{2} \times \dfrac{1}{5} + \dfrac{6}{\left(n + 5\right)\left(n + 4\right)}} = \dfrac{6}{11}$

$⇒ n = 4$