2017 AMC 12B Problems/Problem 20

Problem

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0,1)$. What is the probability that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor$?

$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution

First let us take the case that $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -1$. In this case, both $x$ and $y$ lie in the interval $[{1\over2}, 1)$. The probability of this is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. Similarly, in the case that $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = -2$, $x$ and $y$ lie in the interval $[{1\over4}, {1\over2})$, and the probability is $\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$. It is easy to see that the probabilities for $\lfloor \log_2{x} \rfloor = \lfloor \log_2{y} \rfloor = n$ for $-\infty < n < 0$ are the infinite geometric series that starts at $\frac{1}{4}$ and with common ratio $\frac{1}{4}$. Using the formula for the sum of an infinite geometric series, we get that the probability is $\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{3}}$.

Solution by: vedadehhc

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions

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