Difference between revisions of "2020 CIME II Problems/Problem 6"
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<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 15:42, 6 February 2023
Problem
An infinite number of buckets, labeled ,
,
,
, 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
is
. Given that all three balls land in the same bucket
and that
is even, then the expected value of
can be expressed as
, where
and
are relatively prime positive integers. Find
.
Solution
The probability that all three balls land in box is
. This means that the probability that the three balls land in the same even box is
. This means that the probability that all three balls land in box
is simply
. For events
,
,
,
and probabilities
,
,
,
, the expected value when conducting the experiment is
. Thus, our expected value is just
Our answer is .
~mathboy100