2017 AMC 12B Problems/Problem 20

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Problem 20

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0,1)$. What is the probability that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor$, where $\lfloor r\rfloor$ denotes the greatest integer less than or equal to the real number $r$ ?

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

Solution

$\lfloor\log_2x\rfloor=-1$ when $1/2<x<1$. The chance that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor=-1$ is $1/2*1/2=1/4$. $\lfloor\log_2x\rfloor=-2$ when $1/4<x<1/2$. The chance that $\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor=-2$ is $1/4*1/4=1/16$. This creates an infinite series, where the common ratio is $1/4$. The solution is thus $1/4 * 1/(1-1/4)=D: 1/3$

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