2019 AMC 10A Problem 4

Note that we can pull as many as $14$ balls of each color without ensuring that $15$ balls of one color are drawn. This means that we can draw all of the black, white and blue balls, along with $14$ red, green, and yellow balls. This gives us a total of $$9 + 11 + 13 + 3 \cdot 14$$$$= 33 + 42$$$$= 75.$$ We need to add one at the end, however, to ensure that we get that $15$th ball of some color, $75 + 1 = 76.$ Thus, B is the correct answer.