Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected Six balls are to be randomly chosen from an urn containing $8$ red, $10$ green, and 12 blue balls.
What is the probability at least one red ball, one blue and one green ball is chosen?
Sample space = $\binom{30}{6}$
P = 1 - P(All red + All Green + All Blue + Only red and Green + Only Red and Blue + Only Green and Blue )
$$P = \Large 1 - \frac{\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}}{\binom{30}{6}}$$
According to this, I got $\large 1 - \frac{133099}{593775}$, 
which is $0.7758$?
Is my approach correct?
 A: Your approach is not correct.
In 'Only red and green" you don't exclude situations "Only red" and "only green".
Also each of the situations "Only x" is counted twice - in situation "Only x and y" and "only x and z"
Thus the number of situations where there is a color missing should be:
$$\left(\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}\right)-2\left( \binom{8}{6} + \binom{10}{6} + \binom{12}{6} \right)=$$
$$= \binom{18}{6} + \binom{22}{6} + \binom{20}{6}-\left( \binom{8}{6} + \binom{10}{6} + \binom{12}{6} \right)$$
A: Let $E_1$ be the event that no red ball is chosen, $E_2$ the event that no green ball is chosen, and $E_3$ the event that no blue ball is chosen. 
The probability that at least ball of each color is chosen is
$1 - P(E_1 \cup E_2 \cup E_3).$
By the inclusion-exclusion principle,
$$
P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3) - P(E_1 \cap E_2) - P(E_1 \cap E_3) - P(E_2 \cap E_3) + P(E_1 \cap E_2 \cap E_3).
$$
And we can see that 
$$
P(E_1) = \frac{\binom{22}{6}}{\binom{30}{6}}, \quad
P(E_2) = \frac{\binom{20}{6}}{\binom{30}{6}}, \quad
P(E_3) = \frac{\binom{18}{6}}{\binom{30}{6}}, \quad
P(E_1 \cap E_2) = \frac{\binom{12}{6}}{\binom{30}{6}}, \quad
P(E_1 \cap E_3) = \frac{\binom{10}{6}}{\binom{30}{6}}, \quad
P(E_2 \cap E_3) = \frac{\binom{8}{6}}{\binom{30}{6}}, \quad
P(E_1 \cap E_2 \cap E_3) = 0.
$$
