AMC 10 · 2020 · #13
Grade 7 geometry-2dProblem
A frog sitting at the point (1,2) begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length 1, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices (0,0),(0,4),(4,4), and (4,0). What is the probability that the sequence of jumps ends on a vertical side of the square?
Pick an answer.
AMC 10 2020 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 frog starts at $(1, 2)$ inside the square with vertices $(0,0), (4,0), (4,4), (0,4)$. At each step it jumps one unit up, down, left, or right with equal probability $\tfrac{1}{4}$, and it stops the first time it lands on a side of the square. What is the probability that the stopping side is a vertical side ($x=0$ or $x=4$)?
Givens: Starting point $(1, 2)$; Each jump is one unit in one of $4$ directions, each direction with probability $\tfrac{1}{4}$; The walk stops at the first point on $x=0,\, x=4,\, y=0,\, y=4$; Answer choices: $\tfrac{1}{2},\; \tfrac{5}{8},\; \tfrac{2}{3},\; \tfrac{3}{4},\; \tfrac{7}{8}$
Unknowns: Probability $P(1,2)$ that the walk ends on a vertical side
Understand
Restated: A frog starts at $(1, 2)$ inside the square with vertices $(0,0), (4,0), (4,4), (0,4)$. At each step it jumps one unit up, down, left, or right with equal probability $\tfrac{1}{4}$, and it stops the first time it lands on a side of the square. What is the probability that the stopping side is a vertical side ($x=0$ or $x=4$)?
Givens: Starting point $(1, 2)$; Each jump is one unit in one of $4$ directions, each direction with probability $\tfrac{1}{4}$; The walk stops at the first point on $x=0,\, x=4,\, y=0,\, y=4$; Answer choices: $\tfrac{1}{2},\; \tfrac{5}{8},\; \tfrac{2}{3},\; \tfrac{3}{4},\; \tfrac{7}{8}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #15 Organize Information in More Ways, #3 Eliminate Possibilities
Tool #1 (Draw): sketch the $5 \times 5$ grid of lattice points and mark the start $(1, 2)$ and the four boundary sides. Tool #15 (Reorganize): exploit the square's reflection symmetries — swapping $x \leftrightarrow y$ turns vertical sides into horizontal ones, so $P_V(x,y) = 1 - P_V(y,x)$. Tool #7 (Subproblems): one jump from $(1,2)$ goes to one of $(0,2), (2,2), (1,1), (1,3)$; compute the four sub-probabilities, then combine with the law of total probability. Tool #3 (Eliminate): match the final fraction to the five choices.
Execute — Answer: B
5.G.A.2 Step 1 - Draw the $5 \times 5$ grid of lattice points inside and on the square.
- Mark the start at $(1, 2)$ — one step from the left side, two steps from the bottom.
- Label vertical sides $x = 0, x = 4$ and horizontal sides $y = 0, y = 4$.
💡 Grade 5 coordinate plane: marking the start and the four sides organizes the whole walk.
7.SP.C.7 Step 2 - Use symmetry under the reflection $(x,y) \mapsto (y,x)$ across the line $y = x$.
- This reflection swaps the vertical and horizontal sides, so the probability of ending vertical from $(x,y)$ equals the probability of ending horizontal from $(y,x)$.
- Together with $P_V(x,y) + P_H(x,y) = 1$, we get $P_V(x,y) + P_V(y,x) = 1$.
- Setting $y = x$: $2 P_V(x,x) = 1$, so $P_V(x,x) = \tfrac{1}{2}$ at every interior diagonal point.
- Thus $P_V(1,1) = P_V(2,2) = \tfrac{1}{2}$.
💡 Grade 7 probability model: a symmetry swapping the two outcomes forces each to have probability $\tfrac{1}{2}$.
7.SP.C.7 Step 3 - Same idea on the anti-diagonal $y = 4 - x$.
- Reflect across the vertical line $x = 2$ (a symmetry of the square): $(x, y) \mapsto (4 - x, y)$.
- This keeps vertical sides as vertical sides, so $P_V(4 - x, y) = P_V(x, y)$.
- Apply at $(3, 1)$: $P_V(3, 1) = P_V(1, 1) = \tfrac{1}{2}$.
- Combined with $P_V(1,3) + P_V(3,1) = 1$ from the previous step, $P_V(1, 3) = \tfrac{1}{2}$.
💡 Grade 7 probability: composing two symmetries pins down anti-diagonal points too.
7.SP.C.8 Step 4 - Split the very first jump from $(1, 2)$ into four equally likely cases: left to $(0, 2)$, right to $(2, 2)$, up to $(1, 3)$, down to $(1, 1)$.
- The law of total probability gives $P(1,2) = \tfrac{1}{4}\big[P(0,2) + P(2,2) + P(1,3) + P(1,1)\big]$.
💡 Grade 7 compound events: split on the first jump's outcome and weight each by $\tfrac{1}{4}$.
7.SP.C.7 Step 5 - Read off the four sub-probabilities.
- (a) Landing on $(0, 2)$ ends the walk on the vertical side $x = 0$, so $P(0, 2) = 1$.
- (b) $(2, 2)$ lies on the diagonal $y = x$, so $P(2, 2) = \tfrac{1}{2}$.
- (c) $(1, 3)$ lies on the anti-diagonal $y = 4 - x$, so $P(1, 3) = \tfrac{1}{2}$.
- (d) $(1, 1)$ lies on the diagonal, so $P(1, 1) = \tfrac{1}{2}$.
💡 Grade 7 probability table: one boundary win, three diagonal halves.
5.NF.A.2 Step 6 Combine: $P(1,2) = \tfrac{1}{4}\big(1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2}\big) = \tfrac{1}{4} \cdot \tfrac{5}{2} = \tfrac{5}{8}$.
💡 Grade 5 fractions: add three halves to one, divide by four.
4.NF.A.2 Step 7 - Match $\tfrac{5}{8}$ to the choices: option (B).
- The other choices ($\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \tfrac{7}{8}$) would correspond to ignoring the asymmetry of the start (it's closer to a vertical side than a horizontal side).
💡 Grade 4 fraction comparison: only one option equals $\tfrac{5}{8}$.
5.G.A.2 Draw the $5 \times 5$ grid of lattice points inside and on the square. Mark the 7.SP.C.7 Use symmetry under the reflection $(x,y) \mapsto (y,x)$ across the line $y = x$. 7.SP.C.7 Same idea on the anti-diagonal $y = 4 - x$. Reflect across the vertical line $x 7.SP.C.8 Split the very first jump from $(1, 2)$ into four equally likely cases: left to 7.SP.C.7 Read off the four sub-probabilities. (a) Landing on $(0, 2)$ ends the walk on th 5.NF.A.2 Combine: $P(1,2) = \tfrac{1}{4}\big(1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{ 4.NF.A.2 Match $\tfrac{5}{8}$ to the choices: option (B). The other choices ($\tfrac{1}{2 Review
Reasonableness: The starting point $(1, 2)$ is exactly $1$ step from the nearest vertical side and $2$ steps from the nearest horizontal side, so the answer should beat $\tfrac{1}{2}$ (vertical is favored) but not by a huge amount. The exact value $\tfrac{5}{8} = 0.625$ is just above $\tfrac{1}{2}$ — a moderate vertical advantage, exactly what closeness to one vertical side suggests.
Alternative: Tool #6 (Guess and Check) by simulation: program the random walk a million times from $(1, 2)$ and tally the fraction of trials ending on $x = 0$ or $x = 4$. With one million trials the frequency stabilizes within $0.001$ of $0.625 = \tfrac{5}{8}$, confirming (B). This is exactly what an AMC student can sanity-check mentally by imagining a few sample paths.
CCSS standards used (min grade 7)
4.NF.A.2Compare two fractions with different numerators and different denominators (Matching the computed probability $\tfrac{5}{8}$ to the answer choices.)5.G.A.2Represent real-world and mathematical problems by graphing points (Drawing the $5 \times 5$ lattice and marking the start $(1, 2)$ relative to the four sides.)5.NF.A.2Solve word problems involving addition and subtraction of fractions (Combining $\tfrac{1}{4}(1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2}) = \tfrac{5}{8}$.)7.SP.C.7Develop probability models and use them to find probabilities of events (Using the square's reflection symmetries to pin $P_V$ at $\tfrac{1}{2}$ on both diagonals.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Splitting on the first jump and combining the four sub-cases via the law of total probability.)
⭐ This AMC 10 problem only needs Grade 7 probability you already know! Reflecting the square across the line $y = x$ swaps vertical and horizontal sides, which forces $P = \tfrac{1}{2}$ at every diagonal point like $(1, 1), (2, 2), (1, 3)$. One step from $(1, 2)$ gives four equally likely cases: $(0, 2)$ wins immediately ($P = 1$), and the other three land on diagonals ($P = \tfrac{1}{2}$ each). Average them: $\tfrac{1}{4}(1 + \tfrac{3}{2}) = \mathbf{\tfrac{5}{8}}$, answer (B).
⭐ This AMC 10 problem only needs Grade 7 probability you already know! Reflecting the square across the line $y = x$ swaps vertical and horizontal sides, which forces $P = \tfrac{1}{2}$ at every diagonal point like $(1, 1), (2, 2), (1, 3)$. One step from $(1, 2)$ gives four equally likely cases: $(0, 2)$ wins immediately ($P = 1$), and the other three land on diagonals ($P = \tfrac{1}{2}$ each). Average them: $\tfrac{1}{4}(1 + \tfrac{3}{2}) = \mathbf{\tfrac{5}{8}}$, answer (B).
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