AMC 8 · 2013 · #8

Grade 7 probabilitycounting

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsystematic-enumeration systematic-enumerationtree-enumeration ↑ Prerequisites: fraction-arithmeticsystematic-enumeration

📏 Medium solution 💡 2 insights

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Problem

Toss a fair coin 3 times in a row. What is the probability that somewhere in the three tosses there are at least two heads back-to-back (i.e., the sequence contains "HH")?

Pick an answer.

(A)

$frac{1}{8}$

(B)

$frac{1}{4}$

(C)

$frac{3}{8}$

(D)

$frac{1}{2}$

(E)

$frac{3}{4}$

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

With only 2³ = 8 outcomes, the whole sample space fits on one line — Tool #2 (Systematic List) is the cleanest move: list every 3-toss sequence in a fixed order, then circle the ones that contain HH. Tool #16 (Complement) is held in reserve as a sanity check in Review: counting the outcomes that do NOT contain HH is also short, and the two counts must add to 8.

1STEP 1

Each toss is H or T and the three tosses are independent, so the sample space has 8 equally likely outcomes.

2 × 2 × 2 = 2³ = 8 outcomes

2STEP 2

List every sequence in a fixed order: read each as a binary number (T = 0, H = 1) and count from 000 up to 111 so none is missed.

TTT, TTH, THT, THH, HTT, HTH, HHT, HHH

3STEP 3

Scan each string and mark it a hit if two H's sit next to each other, otherwise a miss.

TTT → miss, TTH → miss, THT → miss, THH → hit, HTT → miss, HTH → miss, HHT → hit, HHH → hit

4STEP 4

The hits are THH, HHT, HHH — 3 favorable out of 8 equally likely sequences, so the probability is $\frac{3}{8}$ , choice (C).

P(at least two consecutive H) = hits/total =

\frac{3}{8}

→ (C)

Answer

\frac{3}{8}

\frac{3}{8}

sits between

\frac{1}{4}

and

\frac{1}{2}

, which feels right: HH somewhere is more likely than getting the exact sequence HHH (probability

\frac{1}{8}

) but less likely than just "two or more heads total" (which would include HTH and gives

\frac{1}{2}

). The answer is also clearly less than

\frac{1}{2}

because of the 8 outcomes, half (4) have at most one H and so cannot contain HH at all.

💡Key takeaway

When the sample space is small, just list every outcome neatly — Grade 7 probability is mostly careful counting!

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