2019 AMC 10A Problem 22

Each chosen number is one of three types: it equals $0$ with probability $\frac14$, equals $1$ with probability $\frac14$, or is uniform on $[0,1]$ with probability $\frac12$. Condition on the types of $x$ and $y$.

Both discrete (probability $\frac12 \cdot \frac12 = \frac14$). Then $|x - y| > \frac12$ exactly when one is $0$ and the other is $1$, which happens with probability $\frac12$ (the four equally likely pairs $00, 01, 10, 11$, two of which work). Contribution: $\frac14 \cdot \frac12 = \frac18$.

One discrete, one uniform (probability $2 \cdot \frac12 \cdot \frac12 = \frac12$). If the discrete value is $0$, the uniform one must exceed $\frac12$; if it is $1$, the uniform one must be below $\frac12$. Either way the chance is $\frac12$. Contribution: $\frac12 \cdot \frac12 = \frac14$.

Both uniform (probability $\frac14$). In the unit square, the region $|x - y| > \frac12$ consists of two right triangles with legs $\frac12$, total area $2 \cdot \frac18 = \frac14$. Contribution: $\frac14 \cdot \frac14 = \frac1{16}$.

Total: $\frac18 + \frac14 + \frac1{16} = \frac{2 + 4 + 1}{16} = \frac{7}{16}$.

The answer is $\boxed{\textbf{(B)}\ \frac{7}{16}}$.

A mixed distribution (point masses plus a continuous part) is handled by conditioning on which part produced each number: the law of total probability turns the problem into a few pure cases, each of which is either a finite count or a single geometric probability. Keep a table of case probability times conditional probability so nothing is double-counted.

• Treating each number as uniform on [0,1] with probability 1/2 and 'a coin value' otherwise, but forgetting that the 0 and 1 outcomes each have probability only 1/4.
• Using $\frac12$ for the both-uniform case (the probability that $|x - y| < \frac12$ is $\frac34$, so the complement is $\frac14$), or counting the mixed case only once instead of twice.