1. Define the events

Let:

• $A$ = the chosen bag is Bag A
• $B$ = the chosen bag is Bag B
• $E$ = the two drawn balls are one red and one black

Since one bag is chosen at random, $P(A)=P(B)=\frac12.$

We are given: $P(A\mid E)=\frac{6}{11}.$

We will use Bayes' theorem: $P(A\mid E)=\frac{P(A)P(E\mid A)}{P(A)P(E\mid A)+P(B)P(E\mid B)}.$

Since $P(A)=P(B)=\tfrac12$, this becomes $\frac{P(E\mid A)}{P(E\mid A)+P(E\mid B)}=\frac{6}{11}.$

2. Find $P(E\mid A)$

Bag A contains:

• $2$ white
• $1$ black
• $3$ red

Total balls in Bag A = $6$.

We want the probability of drawing exactly one red and one black when 2 balls are drawn.

Number of favorable ways: $\binom31\binom11=3.$

Total ways to draw 2 balls from 6: $\binom62=15.$

So, $P(E\mid A)=\frac{3}{15}=\frac15.$

3. Find $P(E\mid B)$

Bag B contains:

• $3$ black
• $2$ red
• $n$ white

Total balls in Bag B = $n+5$.

Favorable ways for one red and one black: $\binom21\binom31=2\cdot 3=6.$

Total ways to draw 2 balls: $\binom{n+5}{2}=\frac{(n+5)(n+4)}{2}.$

Thus, P(E\mid B)=\frac{6}{\binom{n+5}{2}}=rac{12}{(n+5)(n+4)}.

4. Apply the given conditional probability

We have $\frac{P(E\mid A)}{P(E\mid A)+P(E\mid B)}=\frac{6}{11}.$

Substitute the values: $\frac{\frac15}{\frac15+\frac{12}{(n+5)(n+4)}}=\frac{6}{11}.$

Cross-multiply: $11\cdot \frac15 = 6\left(\frac15+\frac{12}{(n+5)(n+4)}\right).$

So, $\frac{11}{5}=\frac{6}{5}+\frac{72}{(n+5)(n+4)}.$

Hence, $\frac{5}{5}=\frac{72}{(n+5)(n+4)}$ $1=\frac{72}{(n+5)(n+4)}.$

Therefore, $(n+5)(n+4)=72.$

Expand: $n^2+9n+20=72$ $n^2+9n-52=0.$

Factorize: $(n+13)(n-4)=0.$

So, $n=-13 \quad \text{or} \quad n=4.$

Since the number of white balls cannot be negative, $n=4.$

5. Check with options

The correct option is: $\boxed{4}$ which is Option C.