AMC 10 · 2020 · #13

Grade 7 geometry-2d

probability-basicsymmetry-argumentcoordinate-geometryrecursive-sequencesystems-of-equations symmetry-argumentcaseworkidentify-subproblems ↑ Prerequisites: probability-basiccoordinate-geometry

📏 Long solution 💡 3 insights

Problem

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.

(A)

$\frac12$

(B)

$\frac 58$

(C)

$\frac 23$

(D)

$\frac34$

(E)

$\frac 78$

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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

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

#1 Draw a Diagram 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 $y = 0, y = 4$.

$$\text{start} = (1, 2),\quad \text{vertical sides: } x = 0,\; x = 4$$

💡 Grade 5 coordinate plane: marking the start and the four sides organizes the whole walk.

#15 Organize Information in More Ways 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 $P_V(1,1) = P_V(2,2) = \tfrac{1}{2}$.

$$P_V(x,x) = \tfrac{1}{2} \;\Rightarrow\; P_V(1,1) = P_V(2,2) = \tfrac{1}{2}$$

💡 Grade 7 probability model: a symmetry swapping the two outcomes forces each to have probability $\tfrac{1}{2}$.

#15 Organize Information in More Ways 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, $P_V(1, 3) = \tfrac{1}{2}$.

$$P_V(1, 3) = 1 - P_V(3, 1) = 1 - \tfrac{1}{2} = \tfrac{1}{2}$$

💡 Grade 7 probability: composing two symmetries pins down anti-diagonal points too.

#7 Identify Subproblems 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 $P(1,2) = \tfrac{1}{4}\big[P(0,2) + P(2,2) + P(1,3) + P(1,1)\big]$.

$$P(1,2) = \tfrac{1}{4}\big[P(0,2) + P(2,2) + P(1,3) + P(1,1)\big]$$

💡 Grade 7 compound events: split on the first jump's outcome and weight each by $\tfrac{1}{4}$.

#7 Identify Subproblems 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 $P(1, 1) = \tfrac{1}{2}$.

$$P(0,2) = 1,\;\; P(2,2) = P(1,3) = P(1,1) = \tfrac{1}{2}$$

💡 Grade 7 probability table: one boundary win, three diagonal halves.

#7 Identify Subproblems 5.NF.A.2 Step 6

Combine: $P(1,2) = \tfrac{1}{4}\big(1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2}\big) = \tfrac{1}{4} \cdot \tfrac{5}{2} = \tfrac{5}{8}$.

$$P(1,2) = \tfrac{1}{4} \cdot \tfrac{5}{2} = \tfrac{5}{8}$$

💡 Grade 5 fractions: add three halves to one, divide by four.

#3 Eliminate Possibilities 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).

$$\tfrac{5}{8} \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 4 fraction comparison: only one option equals $\tfrac{5}{8}$.

[1] #1 5.G.A.2 Draw the $5 \times 5$ grid of lattice points inside and on the square. Mark the

[2] #15 7.SP.C.7 Use symmetry under the reflection $(x,y) \mapsto (y,x)$ across the line $y = x$.

[3] #15 7.SP.C.7 Same idea on the anti-diagonal $y = 4 - x$. Reflect across the vertical line $x

[4] #7 7.SP.C.8 Split the very first jump from $(1, 2)$ into four equally likely cases: left to

[5] #7 7.SP.C.7 Read off the four sub-probabilities. (a) Landing on $(0, 2)$ ends the walk on th

[6] #7 5.NF.A.2 Combine: $P(1,2) = \tfrac{1}{4}\big(1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{

[7] #3 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.2 Compare two fractions with different numerators and different denominators (Matching the computed probability $\tfrac{5}{8}$ to the answer choices.)
5.G.A.2 Represent 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.2 Solve 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.7 Develop 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.8 Find 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).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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