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?

## Best Answer

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 $$ P(E_1) = \frac{\binom{22}{6}}{\binom{30}{6}}, \quad P(E_2) = \frac{\binom{20}{6}}{\binom{30}{6}}, \quad P(E_3) = \frac{\binom{18}{6}}{\binom{30}{6}}, \quad P(E_1 \cap E_2) = \frac{\binom{12}{6}}{\binom{30}{6}}, \quad P(E_1 \cap E_3) = \frac{\binom{10}{6}}{\binom{30}{6}}, \quad P(E_2 \cap E_3) = \frac{\binom{8}{6}}{\binom{30}{6}}, \quad P(E_1 \cap E_2 \cap E_3) = 0. $$