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Difference between revisions of "2020 CIME II Problems/Problem 6"

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==Problem==
 
==Problem==
 
An infinite number of buckets, labeled <math>1</math>, <math>2</math>, <math>3</math>, <math>\ldots</math>, 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 <math>k</math> is <math>2^{-k}</math>. Given that all three balls land in the same bucket <math>B</math> and that <math>B</math> is even, then the expected value of <math>B</math> can be expressed as <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
An infinite number of buckets, labeled <math>1</math>, <math>2</math>, <math>3</math>, <math>\ldots</math>, 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 <math>k</math> is <math>2^{-k}</math>. Given that all three balls land in the same bucket <math>B</math> and that <math>B</math> is even, then the expected value of <math>B</math> can be expressed as <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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==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> <math>\emph{given that they land in the same even box}</math> is simply <math>\frac{63}{64^n}</math>. For events <math>a</math>, <math>b</math>, <math>c</math>, <math>\ldots</math> and probabilities <math>P_a</math>, <math>P_b</math>, <math>P_c</math>, <math>\ldots</math>, the expected value when conducting the experiment is <math>P_aa + P_bb + P_cc + \ldots</math>. Thus, our expected value is just
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<cmath>\frac{2 \cdot 63}{64} + \frac{4 \cdot 63}{64^2} + \frac{6 \cdot 63}{64^3} + \ldots</cmath>
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<cmath> = 2\left(\frac{63}{64} + \frac{63}{64^2} + \ldots\right) + 2\left(\frac{63}{64^2} + \frac{63}{64^3}\right) + \ldots</cmath>
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<cmath> = 2 \cdot 1 + 2 \cdot \frac{1}{64} + 2 \cdot \frac{1}{64^2} + \ldots</cmath>
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<cmath> = 2 \cdot \frac{64}{63}</cmath>
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<cmath> = \frac{128}{63}.</cmath>
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Our answer is <math>128 + 63 = \boxed{191}</math>.
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~mathboy100

Latest revision as of 14:42, 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 $2n$ $\emph{given that they land in the same even box}$ is simply $\frac{63}{64^n}$. For events $a$, $b$, $c$, $\ldots$ and probabilities $P_a$, $P_b$, $P_c$, $\ldots$, the expected value when conducting the experiment is $P_aa + P_bb + P_cc + \ldots$. Thus, our expected value is just

\[\frac{2 \cdot 63}{64} + \frac{4 \cdot 63}{64^2} + \frac{6 \cdot 63}{64^3} + \ldots\] \[= 2\left(\frac{63}{64} + \frac{63}{64^2} + \ldots\right) + 2\left(\frac{63}{64^2} + \frac{63}{64^3}\right) + \ldots\] \[= 2 \cdot 1 + 2 \cdot \frac{1}{64} + 2 \cdot \frac{1}{64^2} + \ldots\] \[= 2 \cdot \frac{64}{63}\] \[= \frac{128}{63}.\]

Our answer is $128 + 63 = \boxed{191}$.

~mathboy100

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