smc

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 $[\frac{1}{2},1)$. 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 $[\frac{1}{4},\frac{1}{2})$, and the probability is $\frac{1}{4}\cdot \frac{1}{4}=\frac{1}{16}$. Recall that the probability that $A$ or $B$ is the case, where case $A$ and case $B$ are mutually exclusive, is the sum of each individual probability. Symbolically that's $P(A\text{ or }B\text{ or }C...)=P(A)+P(B)+P(C)...$. Thus, the probability we are looking for is the sum of the probability for each of the cases $\lfloor {\log }_{2}x\rfloor =\lfloor {\log }_{2}y\rfloor =-1,-2,-\mathrm{3...}$. 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 $\frac{\frac{1}{4}}{1-\frac{1}{4}}=\boxed{\frac{1}{3}}$. Solution by: vedadehhc \\ Edited by: jingwei325