AMC 10 · 2015 · #22

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

probability-basiccomplementary-countinglucas-numbersrecursive-sequence casework ↑ Prerequisites: probability-basic

📏 Long solution 💡 4 insights

Problem

Eight people sit around a circular table and each flips a fair coin. Anyone who gets heads stands up; anyone who gets tails stays seated. Find the probability that no two people who are next to each other are both standing.

Pick an answer.

(A)

$\dfrac{47}{256}$

(B)

$\dfrac{3}{16}$

(C)

$\dfrac{49}{256}$

(D)

$\dfrac{25}{128}$

(E)

$\dfrac{51}{256}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Because every coin is fair and independent, all 2⁸ = 256 stand/sit patterns are equally likely, so the probability is just (number of safe patterns)/256. The phrase "no two adjacent standing" is hard to count all at once, so Tool #7 (Identify Subproblems) splits the work by how many people stand: 0, 1, 2, 3, or 4 (five or more cannot fit without touching). The easy counts come from Tool #2 (Make a Systematic List), and the one stubborn case — exactly 3 standing — bends to Tool #16 (Count the Complement): count all triples, then throw out the ones that put two standers side by side.

1STEP 1

Set up probability as a fraction

8 independent fair coin flips give 2⁸ = 256 equally likely patterns; probability is safe patterns ÷ 256.

P(safe) = (# safe patterns)/2⁸ = (# safe patterns)/256

2STEP 2

Split by how many stand

Sort safe patterns by number of standers; at most 4 can stand (alternating), so only 0-4 standers are possible.

# safe = N₀ + N₁ + N₂ + N₃ + N₄

3STEP 3

Count 0, 1, and 2 standers

Count the small cases directly: 0 standers → 1, 1 stander → 8, 2 standers → C(8, 2) minus 8 neighbor-pairs → 20.

N₀ = 1, N₁ = 8, N₂ = C(8, 2) - 8 = 28 - 8 = 20

4STEP 4

Count 3 standers by complement

Count all triples then subtract bad ones: C(8, 3) − 8 (blocks) − 32 (one pair) = 16 safe patterns.

N₃ = C(8, 3) - 8 - (8× 4) = 56 - 8 - 32 = 16

5STEP 5

Count 4 standers

With 4 sitters left, standers must alternate exactly, giving 2 patterns (even or odd seats).

N₄ = 2 (the two alternating patterns)

6STEP 6

Add up and divide

Adding all cases (1 + 8 + 20 + 16 + 2 = 47) and dividing by 256 gives the answer, choice (A).

P = (1+8+20+16+2)/256 = 47/256 → (A)

Answer

47/256

The probability 47/256≈ 0.18 is sensible: standing is a coin flip, but the no-touching rule rules out most patterns, so a value well under one-half is expected. The counts also pass a sanity check — the safe-pattern totals 1,8,20,16,2 are the Lucas number L₈ = 47, the known count of ways to choose a non-adjacent subset of a circle of 8. The numerator 47 is odd, so the fraction cannot reduce to a denominator like 16 or 128, which quietly rules out (B) 3/16 and (D) 25/128; the nearby (C) 49/256 and (E) 51/256 would require miscounting case 3 or 4.

💡Key takeaway

Every coin flip is equally likely, so count the safe seatings by how many people stand (0-4), add to 47, and divide by 256.

Set up probability as a fraction
Split by how many stand
Count 0, 1, and 2 standers
Count 3 standers by complement
Count 4 standers
Add up and divide