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Difference between revisions of "2020 CIME II Problems/Problem 6"
(→‎Solution)
 
Line 6: Line 6:
  
 
<cmath>\frac{2 \cdot 63}{64} + \frac{4 \cdot 63}{64^2} + \frac{6 \cdot 63}{64^3} + \ldots</cmath>
 
<cmath>\frac{2 \cdot 63}{64} + \frac{4 \cdot 63}{64^2} + \frac{6 \cdot 63}{64^3} + \ldots</cmath>
−
<cmath> = 2(\frac{63}{64} + \frac{63}{64^2} + \ldots) + 2(\frac{63}{64^2} + \frac{63}{64^3}) + \ldots</cmath>
+
<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>
 
<cmath> = 2 \cdot 1 + 2 \cdot \frac{1}{64} + 2 \cdot \frac{1}{64^2} + \ldots</cmath>
 
<cmath> = 2 \cdot 1 + 2 \cdot \frac{1}{64} + 2 \cdot \frac{1}{64^2} + \ldots</cmath>
 
<cmath> = 2 \cdot \frac{64}{63}</cmath>
 
<cmath> = 2 \cdot \frac{64}{63}</cmath>

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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