Let's say we have $30$ balls. $15$ of them are black and the other $15$ remaining are white. We distribute those $30$ balls to $3$ kids, giving each kid $10$ balls.
(A) Taking as a fact that the second kid got $5$ black balls, what's the probability of the first kid to get $5$ black balls?
(B) Taking as a fact that first kids got $10$ black balls, what is the probability of the second kid to get the other $5$ black balls?
What i have thought: Let $B$ be the probability of the second kid to get 5 black balls, $P(B) = \dfrac{\binom{15}{5} \cdot \binom{15}{5}}{\binom{30}{10}}$ because there are $\binom{15}{5}$ ways to get 5 of 15 black balls and $\binom{15}{5}$ to get other 5 balls from the white balls and generally there are $\binom{30}{10}$ ways to get 10 balls from 30. Let $C$ be the probability of the first kid to get 10 black balls from 15 ones, so $P(C) = \dfrac{\binom{15}{10}}{\binom{30}{10}}$. And then, for (A) i need to calculate $P(A \mid B)$. Then for (B) i need to calculate $P(B \mid A)$.
Your question can easily be solved by using the given facts. For example, for the first part, you are given that the second kid has gotten $5$ black and $5$ white balls. As such, we are left with $10$ black and $10$ white balls. So finding the solution of the first question is equivalent to solving the following question:
Can you solve the above problem?
For the second question, can you come with an equivalent problem to be solved?