2019 AMC 10A Problem 22

Step 1: what kind of number can come out?

• First flip heads ($\frac12$), second heads ($\frac12$): the number is $0$. Chance $\frac12 \cdot \frac12 = \frac14$.
• First flip heads, second tails: the number is $1$. Chance $\frac14$.
• First flip tails: the number is a random point anywhere from $0$ to $1$. Chance $\frac12$. Call this "spread out".

Step 2: split into cases by what kind $x$ and $y$ are. For each case, multiply (chance of the case) by (chance that $|x - y| > \frac12$ in that case).

Case A: both are $0$ or $1$. Chance $\frac12 \cdot \frac12 = \frac14$.
In this case the pairs $(0,0), (0,1), (1,0), (1,1)$ are equally likely. The distance is more than $\frac12$ only for $(0,1)$ and $(1,0)$: $2$ out of $4$, so $\frac12$.
Contribution: $\frac14 \cdot \frac12 = \frac18$.

Case B: one is $0$ or $1$, the other is spread out. Chance $\frac12 \cdot \frac12 + \frac12 \cdot \frac12 = \frac12$ (either $x$ or $y$ can be the spread-out one).
If the fixed one is $0$, the spread-out one must be above $\frac12$: chance $\frac12$. If the fixed one is $1$, the spread-out one must be below $\frac12$: chance $\frac12$. Either way, $\frac12$.
Contribution: $\frac12 \cdot \frac12 = \frac14$.

Case C: both are spread out. Chance $\frac12 \cdot \frac12 = \frac14$.
Picture $x$ across and $y$ up in a $1$ by $1$ square. The points with $|x - y| > \frac12$ fill two corner triangles: one below the line $y = x - \frac12$ and one above the line $y = x + \frac12$. Each is a right triangle with legs $\frac12$ and $\frac12$, area $\frac18$. Together: $\frac14$ of the square.
Contribution: $\frac14 \cdot \frac14 = \frac1{16}$.

Step 3: add the cases.
$$\frac18 + \frac14 + \frac1{16} = \frac{2}{16} + \frac{4}{16} + \frac{1}{16} = \frac{7}{16}.$$

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

When a number can come out in different ways, split into cases by which way it came out. Each case is then simple: either a small list of equally likely outcomes, or an area in a square. Multiplying each case's chance by the chance of success inside it, and adding, keeps everything counted exactly once. Check that the case chances add to $1$: $\frac14 + \frac12 + \frac14 = 1$.

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