From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. if the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?

There are y + 12 total balls in the bag. The probability of drawing a blue one is is 12/(y + 12), and the probability of drawing a yellow one is y/(y + 12). We're told that the probability of drawing a blue ball is less than 2/5, so

which tells us there must be at least 19 yellow balls in the bag.

simple 80 pieces in 1 sercing so 80*47=3760+20=3780