Multiplication Rule

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

When a pair of dice is thrown, the ways in which we can get a sum of $$5$$ are: $$(1,4)$$, $$(4,1)$$, $$(2,3)$$ and $$(3,2)$$, giving the total probability among all the $$6\times 6$$ possibilities as $$\dfrac{4}{36}$$

When a pair of dice is thrown, the ways in which we can get a sum of $$8$$ are: $$(2,6)$$, $$(6,2)$$, $$(5,3)$$, $$(3,5)$$, and $$(4,4)$$, giving the total probability among all the $$6\times 6$$ possibilities as $$\dfrac{5}{36}$$

The probability that A wins in the very first throw is $$\dfrac{4}{36}$$

The probability that A wins in the second time he throws is $$\dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{4}{36}$$

The probability that A wins in the third time he throws is $$\dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{4}{36}$$

And so on,

The cases form a GP with the first term $$\dfrac{4}{36}$$ and the common ratio of $$\dfrac{36-4}{36}\times \dfrac{36-5}{36}$$

Thus, the combined probability will be;

$$\dfrac{4/36}{1- \left(\dfrac{36-4}{36}\times \dfrac{36-5}{36}\right)} = \dfrac{144}{304} = \dfrac{36}{76} = \dfrac{9}{19}$$

Option B is the correct answer.