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?
 A: Alfred's answer is correct, though tedious.  The other answers are overcounting, so here I hope to show at least one alternative solution which is slightly less tedious (but still frustrating).  Let us for the time being think of each ball as having a unique label.
Via inclusion-exclusion:
All colors present = total irregardless - at least one color not present + at least two colors not present
(note, "at least 3 or more colors not present" is impossible)
Using symbols and shorthand, let $N_{b}$ represent the event "no blues" and let $N_{b,w}$ represent the event "no blues or whites", etc... and $A$ represent the event "all colors present" and $T$ represent the sure event, you get:
$$A = T - N_b - N_w - N_r + N_{b,w} + N_{b,r} + N_{w,r}$$
Total number of ways of picking 6 balls is then $\binom{17}{6}$
$N_b$ is the number of ways of picking 6 balls in any way so long as there are no blues (even if they are all white or all red or a mix of white and red) for a total of $\binom{17-8}{6}$
$N_w$ is the number of ways of picking 6 balls in any way so long as there are no whites (even if they are all blue or red or a mix of blue and red) for a total of $\binom{17-5}{6}$
Similarly you get $N_r = \binom{17-4}{6}$
You then calculate $N_{b,w}$ is the number of ways of selecting six balls so long as none of them are blue or white, i.e. only red balls.  This would have been $\binom{17-8-5}{6}$ however, notice that $17-8-5=9-5=4$ and $\binom{4}{6}=0$
Similarly $N_{b,r} = \binom{17-8-4}{6}=0$
$N_{w,r} = \binom{17-5-4}{6}=\binom{8}{6}$
So, our total number of cases is:
$$\binom{17}{6}-\binom{17-8}{6}-\binom{17-5}{6}-\binom{17-4}{6}+\binom{17-5-4}{6}$$
$$=\binom{17}{6}-\binom{9}{6}-\binom{12}{6}-\binom{13}{6}+\binom{8}{6}$$
Dividing by the total number of cases, $T=\binom{17}{6}$, gives the probability, $\frac{1210}{1547}\approx 0.78$ wolfram
A: 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!
