Probability with balls and a box - complementary event of "exactly" There are $12$ balls in a box: $b$ blue, $y$ yellow and $3$ red balls. Three balls are randomly chosen. If the probability of choosing one blue, one yellow and one red is $3/11$, find the number of yellow balls in a box.
Attempt:
If $t$ is the total number of balls in a box, then:
$t=3+b+y$
$A$: "We choose exactly one blue, yellow and red ball."
$$P(A)=\frac{b}{t}\cdot\frac{y}{t-1}\cdot\frac{3}{t-2}=\frac{3}{11}$$
Substituting $t$ gives
$$P(A)=\frac{b}{3+b+y}\cdot\frac{y}{2+b+y}\cdot\frac{3}{1+b+y}=\frac{3}{11}$$ 
Now we have one equation with two unknowns, so we need to define another event.
Because we already know $P(A)$, that new event should be the complementary event of $P(A)$.
How to define that event and how to evaluate it?

Remember that you are given $t=12$.
Also,
$$\frac{b}{12}*\frac{y}{11}*\frac{3}{10}=\frac{3}{11}$$ is not correct, as the order in which you draw the balls doesn't matter. There are 6 total ways to draw three distinctly colored balls, so your equation should be:
$$6*\frac{b}{12}*\frac{y}{11}*\frac{3}{10}=\frac{3}{11}$$
This leads to:
$$by=20$$
Now, since we also know that $b+y=9$, we can conclude there are either 4 or 5 yellow balls.
