# 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>[1/2, 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>[\frac{1}{4}, \frac{1}{2})</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>{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>[1/2, 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>[\frac{1}{4}, \frac{1}{2})</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>{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>. |

## Revision as of 21:55, 16 February 2017

## 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 .