<s>[INST] 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?
 [/INST] We have,
b + y + 3 = 12
b + y = 9
b = 9 - y
Probability = $\frac{\binom{9 - y}1 \times \binom{y}1 \times \binom{3}1}{\binom{12}3}$
$\frac 3{11} = \frac{\frac{(9 - y)!}{(8 - y)!.1!} \times \frac{(y)!}{(y - 1)!.1!} \times \frac{3!}{2!.1!}}{\frac{12!}{3!.9!}}$
$\frac 3{11} = \frac{\frac{(9 - y)(8 - y)!}{(8 - y)!} \times \frac{(y)(y - 1)!}{(y - 1)!} \times 3}{2.11.10}$
$\frac 3{11} = \frac{(9 - y) \times (y) \times 3}{2.11.10}$
$20 = (9 - y)(y)$
$ y^2 - 9y + 20 = 0$
On solving yellow y = 4 or y = 5.
Then blue b = 5 or b = 4.
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