AMC 10 · 2019 · #22
Grade 7 probabilityProblem
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∣>21?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A number $x$ in $[0, 1]$ is built as follows: flip a fair coin. If heads, flip again — get $0$ on heads, $1$ on tails. If tails, draw $x$ uniformly from $[0, 1]$. Pick $y$ independently the same way. What is the probability that $|x - y| > \tfrac{1}{2}$?
Givens: First flip Heads ($\tfrac{1}{2}$): a second flip decides $x \in \{0, 1\}$ each with probability $\tfrac{1}{2}$; First flip Tails ($\tfrac{1}{2}$): $x$ is uniform on $[0, 1]$; $y$ is generated by the same independent rule; Goal event: $|x - y| > \tfrac{1}{2}$; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{7}{16}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{9}{16}$, (E) $\tfrac{2}{3}$
Unknowns: $P(|x - y| > \tfrac{1}{2})$
Understand
Restated: A number $x$ in $[0, 1]$ is built as follows: flip a fair coin. If heads, flip again — get $0$ on heads, $1$ on tails. If tails, draw $x$ uniformly from $[0, 1]$. Pick $y$ independently the same way. What is the probability that $|x - y| > \tfrac{1}{2}$?
Givens: First flip Heads ($\tfrac{1}{2}$): a second flip decides $x \in \{0, 1\}$ each with probability $\tfrac{1}{2}$; First flip Tails ($\tfrac{1}{2}$): $x$ is uniform on $[0, 1]$; $y$ is generated by the same independent rule; Goal event: $|x - y| > \tfrac{1}{2}$; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{7}{16}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{9}{16}$, (E) $\tfrac{2}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #1 Draw a Diagram, #9 Solve an Easier Related Problem
Tool #7 (Subproblems): split into four cases by (branch for $x$) $\times$ (branch for $y$) — each independent and equally likely with probability $\tfrac{1}{4}$. Tool #2 (Systematic List): enumerate all four cases (Discrete-Discrete, Discrete-Uniform, Uniform-Discrete, Uniform-Uniform). Tool #1 (Diagram): for the Uniform-Uniform case, draw the unit square and shade $|x-y| > \tfrac{1}{2}$ — two right-triangle corners. Tool #9 (Easier Problem): each sub-case is a small problem solvable by counting or by a clean area; combining them is just a weighted sum.
Execute — Answer: B
7.SP.C.8 Step 1 - Split the sample space by which branch produced $x$ and which produced $y$.
- Branch D (Discrete) has probability $\tfrac{1}{2}$, Branch U (Uniform) has probability $\tfrac{1}{2}$.
- The four joint cases — DD, DU, UD, UU — are independent, each with probability $\tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$.
💡 Independence lets us multiply the two branch probabilities.
7.SP.C.8 Step 2 - Case DD ($x, y \in \{0, 1\}$ uniformly).
- $|x-y| > \tfrac{1}{2}$ requires $x \ne y$, i.e.
- $(x, y) \in \{(0, 1), (1, 0)\}$.
- So $P(|x-y| > \tfrac{1}{2} \mid \text{DD}) = \tfrac{2}{4} = \tfrac{1}{2}$.
💡 List all four equally likely outcomes; two of them work.
7.SP.C.7 Step 3 - Case DU ($x \in \{0, 1\}$, $y$ uniform on $[0, 1]$).
- If $x = 0$: need $|0 - y| = y > \tfrac{1}{2}$, an interval of length $\tfrac{1}{2}$.
- If $x = 1$: need $|1 - y| = 1 - y > \tfrac{1}{2}$, i.e.
- $y < \tfrac{1}{2}$, also length $\tfrac{1}{2}$.
- Either way conditional probability is $\tfrac{1}{2}$.
- Averaging: $P(\cdot \mid \text{DU}) = \tfrac{1}{2}$.
💡 Whichever endpoint $x$ lands on, half of the unit interval is more than $\tfrac{1}{2}$ away.
7.SP.C.7 Step 4 Case UD by symmetry (swap $x$ and $y$): $P(\cdot \mid \text{UD}) = \tfrac{1}{2}$.
💡 Same structure as DU with $x, y$ swapped — answer cannot change.
7.G.B.6 Step 5 - Case UU ($x, y$ both uniform on $[0, 1]$).
- Draw the unit square and shade $|x - y| > \tfrac{1}{2}$, i.e.
- $y > x + \tfrac{1}{2}$ or $y < x - \tfrac{1}{2}$.
- These are two right triangles in opposite corners, each with legs $\tfrac{1}{2}$ and area $\tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8}$.
- Total shaded area $= 2 \cdot \tfrac{1}{8} = \tfrac{1}{4}$, and the unit square has area $1$, so $P(\cdot \mid \text{UU}) = \tfrac{1}{4}$.
💡 Geometric probability: favorable area $\div$ total area on the unit square.
5.NF.A.1 Step 6 - Combine using the law of total probability — each case has weight $\tfrac{1}{4}$: $P = \tfrac{1}{4}(\tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{4}) = \tfrac{1}{4} \cdot \tfrac{7}{4} = \tfrac{7}{16}$.
- The answer is (B).
💡 Average four equally weighted case probabilities.
7.SP.C.8 Split the sample space by which branch produced $x$ and which produced $y$. Bran 7.SP.C.8 Case DD ($x, y \in \{0, 1\}$ uniformly). $|x-y| > \tfrac{1}{2}$ requires $x \ne 7.SP.C.7 Case DU ($x \in \{0, 1\}$, $y$ uniform on $[0, 1]$). If $x = 0$: need $|0 - y| = 7.SP.C.7 Case UD by symmetry (swap $x$ and $y$): $P(\cdot \mid \text{UD}) = \tfrac{1}{2}$ 7.G.B.6 Case UU ($x, y$ both uniform on $[0, 1]$). Draw the unit square and shade $|x - 5.NF.A.1 Combine using the law of total probability — each case has weight $\tfrac{1}{4}$ Review
Reasonableness: $\tfrac{7}{16} \approx 0.44$ is less than $\tfrac{1}{2}$. Intuitively if both $x$ and $y$ were uniform we'd get only $\tfrac{1}{4}$, but the discrete branch boosts the chance — when $x$ or $y$ snaps to an endpoint $0$ or $1$, escaping the other by more than $\tfrac{1}{2}$ is easier. So a number between $\tfrac{1}{4}$ and $\tfrac{1}{2}$ is exactly what we expect. Conversion check: $\tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{2+2+2+1}{4} = \tfrac{7}{4}$, and $\tfrac{1}{4} \cdot \tfrac{7}{4} = \tfrac{7}{16}$. Choice (D) $\tfrac{9}{16}$ is the complementary probability $P(|x-y| \le \tfrac{1}{2})$ trap.
Alternative: Tool #16 (Change Focus / Complement): instead of $|x-y| > \tfrac{1}{2}$ compute $P(|x-y| \le \tfrac{1}{2})$ in each case and subtract from $1$ at the end. Case UU directly: the band $|x-y| \le \tfrac{1}{2}$ has area $\tfrac{3}{4}$ (subtract the two corner triangles from $1$). Same final answer; useful when the complement region is simpler to describe.
CCSS standards used (min grade 7)
5.NF.A.1Add and subtract fractions with unlike denominators (Adding $\tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{4}$ via a common denominator to get $\tfrac{7}{4}$, then multiplying by $\tfrac{1}{4}$.)7.SP.C.7Develop probability models and use them to find probabilities of events (Building the probability model for each branch (Discrete vs Uniform) and computing conditional probabilities of the goal event.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Decomposing the joint experiment into four equally likely independent cases (DD, DU, UD, UU) and combining via the law of total probability.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Computing the area $\tfrac{1}{4}$ of two right-triangle corner regions inside the unit square — geometric probability for the UU case.)
⭐ This AMC 10 problem only needs Grade 7 probability and unit-square geometric probability you already know — split into 4 equally likely cases (Discrete-Discrete, Discrete-Uniform, Uniform-Discrete, Uniform-Uniform), get conditional chances $\tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{4}$, then average: $\tfrac{1}{4}(\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{4}) = \tfrac{7}{16}$.
⭐ This AMC 10 problem only needs Grade 7 probability and unit-square geometric probability you already know — split into 4 equally likely cases (Discrete-Discrete, Discrete-Uniform, Uniform-Discrete, Uniform-Uniform), get conditional chances $\tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{4}$, then average: $\tfrac{1}{4}(\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{1}{4}) = \tfrac{7}{16}$.
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