2019 AMC 10A Problem 22

We can case on whether $x$ and $y$ are chosen from the interval or from $0$ and $1.$ Each case has a $\frac{1}{4}$ chance of happening, since they depend on two coin flips. Case $1: x$ and $y$ are either $0$ or $1$ $x$ and $y$ need to be different, which happens with a $\frac{1}{2}$ probability. Case $2: x$ is either $0$ or $1,$ and $y$ is chosen from $[0, 1]$ If $x = 0,$ then $y$ has to be chosen from $\left(\dfrac{1}{2}, 1\right],$ and if $x = 1,$ then $y$ has to be chosen from $\left[0, \dfrac{1}{2}\right).$ This means that $y$ always has a $\frac{1}{2}$ probability of being chosen from the correct interval. Case $3: x$ is chosen from $[0, 1],$ and $y$ is either $0$ or $1$ This has the same probability as case $2$ due to symmetry. Case $4: x$ and $y$ are chosen from $[0, 1]$ We can use geometric probability since we are working with an infinite number of $(x, y)$ pairs. We graph $|x - y| \gt \frac{1}{2}.$ The shaded area covers $\frac{1}{4}$ of the graph, showing that there is a $\frac{1}{4}$ probability of this case working. Adding the four weighted probabilities gives $$\frac14\left(\frac12+\frac12+\frac12+\frac14\right)=\frac7{16}.$$ Thus, B is the correct answer.