2019 AMC 10B Problem 17

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:geometric distribution complementary probability symmetry

Difficulty rating: 1460

Small hint:

First compute the probability that both balls land in the same bin

Big hint:

If they land in different bins, symmetry decides which color is higher

Solution:

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 .