# [Math] Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected

combinatoricsprobability

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?

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