2021 AMC 12B Problem 23

Problem 23 of 25HarderAlgebra Counting & Probability

Three balls 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

i

2^{-i}

for

i=1,

2,

3,

\ldots.

More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is

\dfrac{p}{q},

where

p

and

q

are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins

3,

17,

and

10.

) What is

p+q?

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Solution

Evenly spaced distinct bins form an arithmetic progression

n,n+d,n+2d

with

n,d\ge 1.

The three labels sum to

3(n+d),

so a fixed assignment of balls to these bins has probability

2^{-3(n+d)}.

The three balls can be ordered in

3!=6

ways, so the total probability is

6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}

=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2

=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}.

Since

\gcd(6,49)=1,

we get

p+q=6+49=55.

Thus, the correct answer is A.

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