2019 AMC 10B Problem 17

Given that the two balls were tossed into separate bins, the probability that the ball in the higher-numbered bin is red is $\frac 12.$ Thus we must find $$\frac{P(\text{Balls in different bins})}2$$$$= \frac{1-P(\text{Balls in same bins})}2$$ by complementary counting. The probability that both balls are in bin $k$ is $$2^{-k} \cdot 2^{-k} = 4^{-k}.$$ The probability that they are both in the same bin is therefore $$\sum_{k=1}^\infty 4^{-k}.$$ Using the geometric sequence formula, we get this to be $$\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13.$$ Therefore, our answer is $$\frac{1-\frac 13}2 = \frac 13 .$$ Thus, the answer is C .