Probability that all colors are chosen A box contains $5$ white, $4$ red, and $8$ blue balls.  You randomly select $6$ balls, without replacement, what is the probability that all three colours are present.

Most similar problems ask for the probability that at least one colour is missing. Which happens to be $1 - P(\text{No colour is missing})$, but how does one find the probability that no colour is missing?
I started by subtracting one ball from each of the possibilities, leaving $4$ White, $3$ red, and $7$ balls and having to select $3$ balls. Is this logic correct? How would it be completed using this method, if it is correct?

Look that you want that all the colors are present, so there are many posible cases. Just to name some:


*

*There are 4 white, 1 red, and 1 blue ball 

*There are 3 white, 2 red, and 1 blue ball

*There are 3 white, 1 red, and 2 blue balls


And the list is much bigger, so what you can do is to use ${n\choose x}$, how? By following the first three cases listed you would have by letting MC be the event in which there are balls of the three colors
$$Pr(MC) = \displaystyle\frac{{5\choose 4}{4\choose 1}{8\choose 1}+{5\choose 3}{4\choose 2}{8\choose 1}+{5\choose 3}{4\choose 1}{8\choose 2}+\dots}{5+4+8 \choose 6}$$
Notice that you should list all the possible cases where there must be at least one ball of every color. Be careful to not repeat cases. Good luck!
