AMC 8 · 2013 · #8
Easy mode Grade 7Problem
Imagine flipping a fair coin 3 times in a row, and writing down each result as H (heads) or T (tails). So you get a string of three letters like HTH or HHT.
We want to find the chance that somewhere in your string there are two heads right next to each other. "TTH" does not count (no two heads side by side). "HHT" does count (the first two letters are both H).
What is the probability that your 3-flip result has two heads in a row?
Pick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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")?
Givens: The coin is fair: each toss is independently $H$ or $T$ with probability $\tfrac{1}{2}$; The coin is tossed exactly $3$ times, so each outcome is an ordered length-$3$ string of $H$s and $T$s; "At least two consecutive heads" means the outcome contains the substring $HH$ (so $HHH$, $HHT$, or $THH$ all count); Answer choices: (A) $\tfrac{1}{8}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{3}{8}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{3}{4}$
Unknowns: The probability that the $3$-toss sequence contains two heads in adjacent positions
Understand
Restated: 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")?
Givens: The coin is fair: each toss is independently $H$ or $T$ with probability $\tfrac{1}{2}$; The coin is tossed exactly $3$ times, so each outcome is an ordered length-$3$ string of $H$s and $T$s; "At least two consecutive heads" means the outcome contains the substring $HH$ (so $HHH$, $HHT$, or $THH$ all count); Answer choices: (A) $\tfrac{1}{8}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{3}{8}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{3}{4}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement
With only $2^3 = 8$ 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 $8$.
Execute — Answer: C
7.SP.C.8 Step 1 - Count the total number of equally likely outcomes.
- Each of the $3$ tosses has $2$ possibilities, and the tosses are independent, so the sample space has $2 \times 2 \times 2 = 8$ ordered sequences.
💡 Setting up a uniform sample space for a compound event is exactly the Grade 7 "organized list for compound events" move.
7.SP.C.8 Step 2 - List every outcome in a fixed order so nothing is missed and nothing is double-counted.
- A simple rule: treat each sequence as a $3$-digit binary number with $T = 0$ and $H = 1$, and list from $000$ up to $111$.
💡 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.7 Step 3 - Mark each outcome as a hit (contains $HH$ somewhere) or a miss.
- Scan each string for two adjacent $H$s.
💡 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.7 Step 4 - Count the hits and form the probability.
- The favorable outcomes are $THH, HHT, HHH$, so there are $3$ of them out of $8$ equally likely sequences.
💡 For a uniform sample space, probability is just (favorable count) / (total count) — the basic Grade 7 probability formula.
7.SP.C.8 Count the total number of equally likely outcomes. Each of the $3$ tosses has $2 7.SP.C.8 List every outcome in a fixed order so nothing is missed and nothing is double-c 7.SP.C.7 Mark each outcome as a hit (contains $HH$ somewhere) or a miss. Scan each string 7.SP.C.7 Count the hits and form the probability. The favorable outcomes are $THH, HHT, H Review
Reasonableness: $\tfrac{3}{8}$ sits between $\tfrac{1}{4}$ and $\tfrac{1}{2}$, which feels right: $HH$ somewhere is more likely than getting the exact sequence $HHH$ (probability $\tfrac{1}{8}$) but less likely than just "two or more heads total" (which would include $HTH$ and gives $\tfrac{1}{2}$). The answer is also clearly less than $\tfrac{1}{2}$ because of the $8$ outcomes, half ($4$) have at most one $H$ and so cannot contain $HH$ at all.
Alternative: Tool #16 (Complement) gives a quick cross-check. The outcomes that do NOT contain $HH$ are exactly the strings whose $H$s are separated by at least one $T$: $TTT, TTH, THT, HTT, HTH$ — that is $5$ outcomes. So $P(\text{no } HH) = \tfrac{5}{8}$, and $P(\text{at least one } HH) = 1 - \tfrac{5}{8} = \tfrac{3}{8}$, confirming (C).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Building the full $8$-outcome sample space for $3$ coin tosses and listing every sequence in a fixed order so the favorable cases can be identified.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating each of the $8$ ordered sequences as equally likely and computing the event probability as $\tfrac{\text{favorable count}}{\text{total count}} = \tfrac{3}{8}$.)
⭐ When the sample space is small, just list every outcome neatly — Grade 7 probability is mostly careful counting!
⭐ When the sample space is small, just list every outcome neatly — Grade 7 probability is mostly careful counting!
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