AMC 8 · 2022 · #25

Grade 7 probabilitycounting

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicrecursive-sequencesystematic-enumeration tree-enumerationcaseworkcomplementary-counting ↑ Prerequisites: probability-basicsystematic-enumeration

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A cricket sits on one of 4 leaves. Each turn it jumps to one of the other 3 leaves, chosen uniformly at random. After exactly 4 jumps, what is the probability that it is back on the leaf it started from?

Pick an answer.

(A)

$frac{2}{9}$

(B)

$frac{19}{80}$

(C)

$frac{20}{81}$

(D)

$frac{1}{4}$

(E)

$frac{7}{27}$

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Going from 4 jumps to the answer in one shot is hard, but going from n jumps to n+1 jumps is easy — that is exactly Tool #9 (Solve an Easier Related Problem) applied to the time variable. Use Tool #1 (Draw a Diagram) to picture the 4 leaves with edges between every pair (K₄); the picture makes the symmetry obvious — leaves B, C, D are interchangeable, so they always share the same probability q_n of holding the cricket after n jumps. That symmetry collapses 4 unknowns down to just two (p_n for leaf A and q_n for any other leaf), and Tool #5 (Look for a Pattern) then turns the problem into a tiny one-step recursion that we run for n=1,2,3,4.

1STEP 1

Draw leaves A, B, C, D and join every pair — the graph K₄; each leaf has 3 equally likely jumps, and B, C, D play symmetric roles.

P(jump to any specific other leaf) =

\frac{1}{3}

2STEP 2

Track two numbers after n jumps: p_n = chance on A, q_n = chance on one specific other leaf; symmetry forces p_n + 3 q_n = 1.

p_n + 3 q_n = 1, p₀ = 1, q₀ = 0

3STEP 3

One-step rules: reach A only from an other leaf that picks A, so p_n+1 = q_n; reach a given other by two paths, q_n+1 = $\frac{p_n + 2 q_n}{3}$ .

p_n+1 = q_n, q_n+1 =

\frac{p_n + 2 q_n}{3}

4STEP 4

Run it from p₀ = 1, q₀ = 0: p₁ = 0, p₂ = $\frac{1}{3}$ , p₃ = $\frac{2}{9}$ , then p₄ = $\frac{7}{27}$ , checking p_n + 3 q_n = 1 each step.

n=1:& p₁ = 0, q₁ =

\frac{1}{3}

; n=2:& p₂ =

\frac{1}{3}

, q₂ =

\frac{0 + \frac{2}{3}}{3}

\frac{2}{9}

; n=3:& p₃ =

\frac{2}{9}

, q₃ =

\frac{\frac{1}{3} + \frac{4}{9}}{3}

\frac{\frac{7}{9}}{3}

\frac{7}{27}

; n=4:& p₄ =

\frac{7}{27}

5STEP 5

After 4 jumps the cricket is back on start leaf A with probability $\frac{7}{27}$ , which is choice (E).

p₄ =

\frac{7}{27}

→ (E)

Answer

frac{7}{27}

Sanity check 1: p_n + 3 q_n = 1 at every step (e.g.

\frac{7}{27}

+ 3 ·

\frac{20}{81}

\frac{21}{81}

\frac{60}{81}

= 1). Sanity check 2: if jumps were random over all 4 leaves (including staying), the long-run probability of being on A would be

\frac{1}{4}

= 0.25. Our answer

\frac{7}{27}

≈ 0.259 is just slightly above

\frac{1}{4}

, which makes sense because 4 is an even number of jumps and the cricket is forced to move each turn, very slightly biasing it back toward A. Choices (A)

\frac{2}{9}

≈ 0.222 and (D)

\frac{1}{4}

= 0.25 are too low; (B)

\frac{19}{80}

has the wrong denominator family (should involve 3ⁿ). Only (E) fits.

💡Key takeaway

This AMC 8 problem only needs Grade 7 compound-event probability — "chance of step 1 times chance of step 2 times..." — that you already know!

More like this