Difference between revisions of "2017 AMC 12B Problems/Problem 20"
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− | First let us take the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1</math>. In this case, both <math>x</math> and <math>y</math> lie in the interval <math>[1\over2, 1)</math>. The probability of this is <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>. Similarly, in the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -2</math>, <math>x</math> and <math>y</math> lie in the interval <math>[{1\over4}, {1\over2})</math>, and the probability is <math>\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}</math>. It is easy to see that the probabilities for <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = n</math> for <math>-\infty < n < 0</math> are the infinite geometric series that starts at <math>\frac{1}{4}</math> and with common ratio <math>\frac{1}{4}</math>. Using the formula for the sum of an infinite geometric series, we get that the probability is <math>\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}</math>. | + | First let us take the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1</math>. In this case, both <math>x</math> and <math>y</math> lie in the interval <math>[{1\over2}, 1)</math>. The probability of this is <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>. Similarly, in the case that <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -2</math>, <math>x</math> and <math>y</math> lie in the interval <math>[{1\over4}, {1\over2})</math>, and the probability is <math>\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}</math>. It is easy to see that the probabilities for <math>\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = n</math> for <math>-\infty < n < 0</math> are the infinite geometric series that starts at <math>\frac{1}{4}</math> and with common ratio <math>\frac{1}{4}</math>. Using the formula for the sum of an infinite geometric series, we get that the probability is <math>\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}</math>. |
Solution by: vedadehhc | Solution by: vedadehhc |
Revision as of 21:51, 4 February 2019
Problem 20
Real numbers and
are chosen independently and uniformly at random from the interval
. What is the probability that
, where
denotes the greatest integer less than or equal to the real number
?
Solution
First let us take the case that . In this case, both
and
lie in the interval
. The probability of this is
. Similarly, in the case that
,
and
lie in the interval
, and the probability is
. It is easy to see that the probabilities for
for
are the infinite geometric series that starts at
and with common ratio
. Using the formula for the sum of an infinite geometric series, we get that the probability is
.
Solution by: vedadehhc
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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