There are $150$ balls altogether, and you’re choosing $20$ of them; a set with $150$ members has $\binom{150}{20}$ $20$-element subsets, so there are $\binom{150}{20}$ sets of $20$ balls that you could draw, all of them equally likely to be drawn. That accounts for the denominator of the fraction: it’s the number of equally likely possible outcomes.
Similarly, there are $\binom{40}{10}$ different sets of $10$ white balls, $\binom{50}4$ different sets of $4$ red balls and $\binom{60}6$ different sets of $60$ black balls. There are $$\binom{40}{10}\binom{50}4\binom{60}6$$ ways to combine one of the $\binom{40}{10}$ possible sets of $10$ white balls, one of the $\binom{50}4$ possible sets of $4$ red balls, and one of the $\binom{60}6$ possible sets of $6$ black balls, so there are $$\binom{40}{10}\binom{50}4\binom{60}6$$ successful outcomes, where successful means having $10$ white balls, $4$ red balls, and $6$ black balls. As usual, the probability of a successful outcome is the ratio of successful outcomes to equally likely possible outcomes, or
$$\frac{\binom{40}{10}\binom{50}4\binom{60}6}{\binom{150}{20}}\;.$$
Nothing in the analysis changes in any way if we paint numbers from $1$ through $150$ on the balls to give them unique identities: we still have to count the number of sets of $10$ white balls, the number of sets of $4$ red balls, the number of sets of $6$ black balls, and the number of sets of $20$ balls of whatever colors, and we still have to perform the same calculations with these numbers.
Added: I’m afraid that the calculation in the edit makes little sense. To see this more clearly, let’s look at a simpler example. Suppose that there are $99$ white balls and $1$ red ball, and you draw $20$ balls at random without replacement. By your reasoning there are $\binom{21}1=21$ possible outcomes, ranging from $20$ white and no red balls to no white and $20$ red balls. This, however, is clearly not the case, since there is only $1$ red ball in the bag. Okay, suppose that you take this limitation into account: then by your approach there are exactly two possible outcomes: $20$ white balls, and $19$ white balls and $1$ red balls, so the denominator of your fraction will be $2$.
For the outcome of drawing $20$ white balls the numerator will be $\frac{20!}{20!}=1$, so you would conclude that the probability of getting $20$ white balls is $\frac12$. Is that reasonable? Notice that you’d get the same result if there were $999$ white balls and $1$ red ball, or $999999$ white balls and $1$ red ball. I think that it’s pretty clear that this cannot be right.
Worse, your numerator for the outcome of drawing $19$ white balls and the red ball will be $\frac{20!}{1!19!}=20$, and the ‘probability’ of this outcome will be $\frac{20}2=10$. That certainly can’t be right!
What you’re missing is that even if there is nothing to distinguish one white ball from another, there are still $99$ different white balls in the bag, and a $99$-element set has $\binom{99}{20}$ different $20$-element subsets. Every one of these $\binom{99}{20}$ sets is a different outcome, even if you have no way to tell which of them you actually got. Thus, in the simplified problem the probability of getting $20$ white marbles is actually
$$\frac{\binom{99}{20}}{\binom{100}{20}}=\frac{99!}{20!79!}\cdot\frac{20!80!}{100!}=\frac{80}{100}=\frac45\;,$$
while the probability of getting $19$ white balls and the red ball is simply the probability of getting the red ball, $$\frac{20}{100}=\frac15\;.$$
Let $b_1$ be the event drawing a black ball the first draw; and $b_2$ the event of drawing a black ball the second draw.
We want $P(b_2|b_1)$. That is, we want $\frac{P(b_2\cap b_1)}{P(b_1)}$.
We have $P(b_1)=\frac12\cdot\frac12+\frac12\cdot\frac34=\frac58$.
To figure out $P(b_1\cap b_2)$ we note that there are four ways this can happen:
(i) Bin A, Black, Bin A, Black, with probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac13=\frac1{24}$
(ii) Bin A, Black, Bin B, Black, with probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac34=\frac3{32}$
(iii) Bin B, Black, Bin A, Black, with probability $\frac12\cdot\frac34\cdot\frac12\cdot\frac12=\frac3{32}$
(iv) Bin B, Black, Bin B, Black, with probability $\frac12\cdot\frac34\cdot\frac12\cdot\frac23=\frac18$
So $P(b_1\cap b_2)=\frac{1}{24}+\frac3{32}+\frac3{32}+\frac18=\frac{34}{96}=\frac{17}{48}$
Then the overall conditional probability $P(b_2|b_1)=\frac{\frac{17}{48}}{\frac58}=\frac{17}{30}$
Best Answer
Suppose we instead had six bags: five of them get a black ball each, and the last gets the eight white balls. We first pick a bag at random, then pick a ball from the bag at random. In this case the chance of getting a black ball is 5/6. The reason is that the thirteen balls are not equally likely to get picked -- the five black balls have each as much chance as getting picked as all the white balls combined. Hence a two-step process like this alters the odds from the single bag experiment.
Or, taken to extremes, suppose I tell you that in one of my hands is the winning lottery ticket (odds of winning one in 40 million) and in the other of my hands are all the losing lottery tickets (compressed in some way to fit into my hand). Obviously you would be happy to play this game, as you have a 50-50 chance of getting the winner.