AMC 8 · 2013 · #8

Grade 7 probabilitycounting

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

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Problem

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

Pick an answer.

(A)

$frac{1}{8}$

(B)

$frac{1}{4}$

(C)

$frac{3}{8}$

(D)

$frac{1}{2}$

(E)

$frac{3}{4}$

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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")?

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

#2 Make a Systematic List 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.

$$2 \times 2 \times 2 = 2^3 = 8 \text{ outcomes}$$

💡 Setting up a uniform sample space for a compound event is exactly the Grade 7 "organized list for compound events" move.

#2 Make a Systematic List 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 $111$.

$$TTT, \; TTH, \; THT, \; THH, \; HTT, \; HTH, \; HHT, \; HHH$$

💡 Tool #2's golden rule: pick an ordering and stick to it. Counting from $000$ to $111$ guarantees all $8$ sequences appear exactly once.

#2 Make a Systematic List 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.

$$TTT \to \text{miss}, \; TTH \to \text{miss}, \; THT \to \text{miss}, \; THH \to \textbf{hit}, \; HTT \to \text{miss}, \; HTH \to \text{miss}, \; HHT \to \textbf{hit}, \; HHH \to \textbf{hit}$$

💡 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.

#2 Make a Systematic List 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.

$$P(\text{at least two consecutive } H) = \dfrac{\text{hits}}{\text{total}} = \dfrac{3}{8} \;\Rightarrow\; \textbf{(C)}$$

💡 For a uniform sample space, probability is just (favorable count) / (total count) — the basic Grade 7 probability formula.

[1] #2 7.SP.C.8 Count the total number of equally likely outcomes. Each of the $3$ tosses has $2

[2] #2 7.SP.C.8 List every outcome in a fixed order so nothing is missed and nothing is double-c

[3] #2 7.SP.C.7 Mark each outcome as a hit (contains $HH$ somewhere) or a miss. Scan each string

[4] #2 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.8 Find 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.7 Develop 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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