Competition · AMC preparation · step 4 of 4
AMC 8 · 2013 · #8
Grade 7 probabilitycountingPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count all the outcomes
Each toss is H or T and the three tosses are independent, so the sample space has 8 equally likely outcomes.
Setting up a uniform sample space for a compound event is exactly the Grade 7 "organized list for compound events" move.
7.SP.C.8Make A Systematic ListList every outcome
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.
Tool #2's golden rule: pick an ordering and stick to it. Counting from 000 to 111 guarantees all 8 sequences appear exactly once.
7.SP.C.8Make A Systematic ListMark the hits
Scan each string and mark it a hit if two H's sit next to each other, otherwise a miss.
Defining the event by checking each outcome against the rule is the Grade 7 "develop a probability model" idea — every outcome gets a yes/no label.
7.SP.C.7Make A Systematic ListForm the probability
The hits are THH, HHT, HHH — 3 favorable out of 8 equally likely sequences, so the probability is , choice (C).
For a uniform sample space, probability is just (favorable count) / (total count) — the basic Grade 7 probability formula.
The probability of getting two heads in a row somewhere in three tosses equals the number of equally likely three-toss sequences that contain HH divided by the total number of such sequences.
▸ Why?
All eight sequences are equally likely, so the chance that HH appears is simply how many of the eight sequences contain HH out of all eight — its favorable count over the total count.
▸ Why?
There are eight sequences in the denominator because each of the three tosses independently offers two outcomes, and two options repeated across three tosses give 2 × 2 × 2 = 8 equally likely sequences.
▸ Why?
Each toss doubles the running list — every earlier sequence branches into a Heads version and a Tails version — so three fair tosses stack as three equal rounds of doubling, 2 × 2 × 2.
▸ Why?
Exactly three of the sequences contain two adjacent heads, because writing all eight in the fixed order 000 up to 111 (with T = 0 and H = 1) lists every sequence once, and scanning them finds HH only in THH, HHT, and HHH.
▸ Why?
Pairing each sequence with one binary number from 000 to 111 matches the sequences one-for-one with the eight numbers, so none is skipped and none is repeated, making the three hits a complete tally.
When the sample space is small, just list every outcome neatly — Grade 7 probability is mostly careful counting!
- Count all the outcomes
- List every outcome
- Mark the hits
- Form the probability
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