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2021 AMC 12B Problem 23

Problem 23 of 25HarderAlgebraProbability & Statistics

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 ii is 2−i2^{-i} for i=1,i=1, 2,2, 3,3, ….\ldots. More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is pq,\dfrac{p}{q}, where pp and qq are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins 3,3, 17,17, and 10.10.) What is p+q?p+q?

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Solution

Evenly spaced distinct bins form an arithmetic progression n,n+d,n+2dn,n+d,n+2d with n,d≥1.n,d\ge 1. The three labels sum to 3(n+d),3(n+d), so a fixed assignment of balls to these bins has probability 2−3(n+d).2^{-3(n+d)}. The three balls can be ordered in 3!=63!=6 ways, so the total probability is 6∑n≥1∑d≥12−3(n+d)6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)} =6(∑n≥118n)2=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2 =6⋅17⋅17=649.=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}. Since gcd⁡(6,49)=1,\gcd(6,49)=1, we get p+q=6+49=55.p+q=6+49=55. Thus, the correct answer is A.
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Tagged: basic probability · geometric sequence · arithmetic sequence

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