AMC 8 · 2013 · #8
Grade 7 probabilitycountingProblem
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
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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