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Difference between revisions of "2020 CIME II Problems/Problem 6"
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==Solution==
 
==Solution==
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The probability that all three balls land in box <math>2n</math> is <math>\frac{1}{64^n}</math>. This means that the probability that the three balls land in the same even box is <math>\frac{1}{64} + \frac{1}{64^2} + \ldots = \frac{1}{63}</math>. This means that the probability that all three balls land in box <math>2n</math> \emph{given that they land in the same even box}
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The probability that all three balls land in box <math>2n</math> is <math>\frac{1}{64^n}</math>. This means that the probability that the three balls land in the same even box is <math>\frac{1}{64} + \frac{1}{64^2} + \ldots = \frac{1}{63}</math>. This means that the probability that all three balls land in box <math>2n</math> <math>\emph{given that they land in the same even box}</math>

Revision as of 15:34, 6 February 2023

Problem

An infinite number of buckets, labeled $1$, $2$, $3$, $\ldots$, lie in a line. A red ball, a green ball, and a blue ball are each tossed into a bucket, such that for each ball, the probability the ball lands in bucket $k$ is $2^{-k}$. Given that all three balls land in the same bucket $B$ and that $B$ is even, then the expected value of $B$ can be expressed as $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

The probability that all three balls land in box $2n$ is $\frac{1}{64^n}$. This means that the probability that the three balls land in the same even box is $\frac{1}{64} + \frac{1}{64^2} + \ldots = \frac{1}{63}$. This means that the probability that all three balls land in box $2n$ $\emph{given that they land in the same even box}$

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