1. Let

   • $A$ = event that the first selected ball is black
   • $B$ = event that the second selected ball is black

   We need to find: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

2. Total balls = $4$ white $+ 6$ black $= 10$.

3. Compute $P(A\cap B)$

   This means both first and second balls are black.

   $P(A\cap B)=P(\text{first black and second black})$ $=\frac{6}{10}\cdot\frac{5}{9}=\frac{30}{90}=\frac{1}{3}$

4. Compute $P(B)$

   Probability that the second ball is black.

   We can use symmetry: since the balls are drawn randomly, the probability that the second ball is black is the same as the probability that any particular draw is black.

   There are $6$ black balls out of $10$, so $P(B)=\frac{6}{10}=\frac{3}{5}$

   (Alternatively, by total probability: $P(B)=\frac{6}{10}\cdot\frac{5}{9}+\frac{4}{10}\cdot\frac{6}{9}=\frac{1}{3}+\frac{4}{15}=\frac{3}{5}$)

5. Now apply conditional probability: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3}\cdot\frac{5}{3}=\frac{5}{9}$

6. Thus, $\frac{m}{n}=\frac{5}{9}$ where $\gcd(5,9)=1$.

   So, $m+n=5+9=14$

7. Checking options:

   • A: $4$
   • B: $14$ ✅
   • C: $11$
   • D: $13$

Therefore, the correct answer is Option B.