2019 AMC 10A Problem 22

P21-25~6 minPrint

Conditional Probability & States Geometric Probability Split into exhaustive cases and handle each

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 $|x-y| > \tfrac{1}{2}$ ?

Show solution without answering

Key insight

Each number is 0 (chance 1/4), 1 (chance 1/4) or a random point in between (chance 1/2). Split into cases by type and add: 1/8 + 1/4 + 1/16.

Solution

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

Why this works

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 $1$ : $\frac14 + \frac12 + \frac14 = 1$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Introduction to Counting & Probability, Chapter 10: Geometric Probability
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Intermediate Counting & Probability, Chapter 11: Conditional Probability