2021 AMC 12B Spring Problem 23

Question

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

is

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?

55

56

57

58

59

Answer: A

Concepts:basic probability geometric sequence arithmetic sequence

Difficulty rating: 2390

Small hint:

Evenly spaced means the bins are

n, n+d, n+2d

for some

n,d\ge 1;

their labels sum to

3(n+d)

Big hint:

Each such ordered assignment has probability

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

and there are

3!

orderings of the three balls

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.