probability, at least one is not present

Question

you have $5$ red balls, $10$ green balls, and $15$ yellow balls in a balls. You randomly choose $5$ without replacement.

What is the probability that at least one of the $3$ colors is not present among the $5$ you pick? Hint: $P(A \cup B \cup C)=$?

attempt:

$P(\text{at least one not present})=1-P(\text{all 3 present})$. Then I got stuck. I expand $$ P( A \cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C) \\ + P(A \cap B \cap C). $$ But I got stuck too.

Answer 1

Atleast one of the three colors not present =

$P(E)= 1-P(All) = 1-P(A \cap B \cap C)$

Now use your formula the other way round.