2017 AMC 12B Problem 20

Question

Real numbers

x

and

y

are chosen independently and uniformly at random from the interval

(0, 1).

What is the probability that

\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor,

where

\lfloor r \rfloor

denotes the greatest integer less than or equal to the real number

r?

\dfrac{1}{8}

\dfrac{1}{6}

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

\lfloor \log_2 x \rfloor = -n

exactly when

\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}

For each

n,

both

x

and

y

must land in that interval of length

\dfrac{1}{2^n};

sum the resulting areas

For each positive integer

n,

\lfloor \log_2 x \rfloor = -n

exactly when

\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}},

an interval of length

\dfrac{1}{2^n}.

The event that both floors equal

-n

is a square of area

\dfrac{1}{4^n}.

Summing over all

n,

the probability is

\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}.

Thus, the correct answer is D.

Problem 20 in other years