2019 AMC 12B Problem 13

Question

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin

k

is

2^{-k}

for

k=1,

2,

3,

\ldots

What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

\dfrac{1}{4}

\dfrac{2}{7}

\dfrac{1}{3}

\dfrac{3}{8}

\dfrac{3}{7}

Answer: C

Concepts:basic probability symmetry complementary probability

Difficulty rating: 1440

Small hint:

By symmetry, red-higher and green-higher are equally likely

Big hint:

Find the probability of a tie,

\displaystyle\sum_{k\ge1}\left(2^{-k}\right)^2

Solution:

The probability the balls land in the same bin is

\begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac13. \end{gathered}

By symmetry, the red ball being higher and the green ball being higher are equally likely, so each has probability

\dfrac{1-\tfrac13}{2}=\dfrac13.

Thus, C is the correct answer.