There are 7 red balls, 9 black balls and 3 white balls in a box. What is the chance to draw two black balls?

You didn't mention if the balls drawn with or without replacement.

However I will give you both probabilities

A) Ball drawn and replaced before the second draw:

Sample size = {7 Red, 9 Black, 3 White} = 19 possible outcomes

Favorable outcome={drawing black ball}

P (1st black draw) = 9/19 = 0.4736

P (2nd black draw) = 9/19 = 0.4736

P (1st black draw AND 2nd black draw) = 0.4736 x 0.4736 = 0.224

B) Ball drawn and not replaced:

P (1st black draw) = 9/19 = 0.4736.

Now that one black is drawn and not replaced, then the remaining number of black is now 8 and the total 18 instead of 19:

P (2nd black draw) = 8/18 = 0.4444

P (1st black draw AND 2nd black draw without replacement) =

0.4736. x 0.4444 = 0.21

You have a chance of 1/2 of getting a black ball. Good luck