Determine the probability that all first 3 balls are black, knowing that at least one of them is black. 
A box has 10 balls, 6 of which are black and 4 are white. Three balls are removed randomly from the box. Determine the probability that all first 3 balls are black, knowing that at least one of them is black.

Attempt:
So we need to calculate $P($obtain $3$ black balls$|$ at least $1$ is black$)$. We can separate this probability into three different probabilities:
$$ P(\text{Obtain 3 black balls}|\text{At least 1 is black}) = P(\text{Obtain 3 black balls}|\text{1 is black}) + P(\text{Obtain 3 black balls}|\text{2 are black}) + P(\text{Obtain 3 black balls}|\text{3 are black}) $$
The last probability, I suppose, is equal to $(6/10) \times (5/9) \times (4/8) $. The first two I'm unsure how to calculate.
 A: By definition
$$
P(\text{ $3$ black balls }| \text{ at least $1$ black ball })= \frac{ P(\text{ $3$ black balls and at least $1$ black ball )}}{P(\text{ at least $1$ black ball })}\ .
$$
Now $\ P(\text{ $3$ black balls and at least $1$ black ball })=$$P(\text{ $3$ black balls }) $, so what is $\ P(\text{ $3$ black balls }) $, and what is $\ P(\text{ at least $1$ black ball })\ $?
A: The easy way to solve it is to observe that the solution can be represented as
$$\frac{\mathbb{P}[BBB]}{1-\mathbb{P}[WWW]}=\frac{\frac{\binom{6}{3}}{\binom{10}{3}}}{1-\frac{\binom{4}{3}}{\binom{10}{3}}}=\frac{\binom{6}{3}}{\binom{10}{3}-\binom{4}{3}}=\frac{6\times5\times4}{10\times9\times8-4\times3\times2}=\frac{5}{30-1}=\frac{5}{29}$$
A: The RHS in your effort is the (unconditional) probability to obtain 3 black balls (so is not the same as the LHS).

Let $B$ denote the number of black balls and let $W$ the number of white balls.
Then:
$$P(B=3\mid B\geq1)=P(B=3,B\geq1)/P(B\geq1)=P(B=3)/(1-P(W=3))$$
Here $P(B)=\frac6{10}\frac59\frac48$ and $P(W=3)=\frac4{10}\frac39\frac28$.
