Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $$\frac{m}{n}$$, where $$gcd(m,n)=1$$, then $$m+n$$ is equal to :

We need to find $$ P(\text{first black} \mid \text{second black}) $$

We know, for conditional probability, $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$

Let, $$ A$$ be the event that the first ball is black, and $$ B$$ be that the second ball is black

So, probability both balls are black (without replacement) is $$ P(A \cap B) = \dfrac{6}{10} \cdot \dfrac{5}{9} = \dfrac{1}{3} $$

Probability second ball is black is 

$$ P(B) = P(\text{second black | first black}) \cdot P(\text{first black}) + P(\text{second black | first white}) \cdot P(\text{first white}) $$

$$ = \dfrac{5}{9} \cdot \dfrac{6}{10} + \dfrac{6}{9} \cdot \dfrac{4}{10} = \dfrac{30}{90} + \dfrac{24}{90} = \dfrac{3}{5} $$

Now,  $$ P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1/3}{3/5} = \dfrac{5}{9} $$

Thus, $$ m+n = 5+9 =14$$ .