Question

Real numbers between $0$ and $1,$ inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is $0$ if the second flip is heads and $1$ if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0, 1].$ Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x - y| \gt \dfrac{1}{2}?$

$\dfrac{1}{3}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{2}{3}$

Solution:

Each variable equals $0$ with probability $\tfrac14,$ equals $1$ with probability $\tfrac14,$ and is uniform on $[0, 1]$ with probability $\tfrac12.$

Considering the nine combinations of types: the pairs $(0, 1)$ and $(1, 0)$ each contribute $\tfrac{1}{16}.$ Each of the four point-versus-uniform cases contributes $\tfrac{1}{16}.$ The uniform-versus-uniform case contributes $\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}.$

The total is $\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}.$

Thus, the correct answer is B.

2019 AMC 12A Problem 20

Solution:

Problem 20 in Other Years