AMC 8 · 2002 · #21
Grade 7 probabilitycountingProblem
Harold tosses a coin four times. The probability that he gets at least as many heads as tails is
Pick an answer.
AMC 8 2002 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: Harold tosses a fair coin $4$ times. What is the probability that the number of heads is at least the number of tails?
Givens: A fair coin is tossed $4$ times, so each of the $2^4 = 16$ outcome sequences is equally likely; We want $P(\text{heads} \geq \text{tails})$ across the $4$ tosses; Answer choices: (A) $\tfrac{5}{16}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{5}{8}$, (E) $\tfrac{11}{16}$
Unknowns: The probability that heads $\geq$ tails in $4$ tosses
Understand
Restated: Harold tosses a fair coin $4$ times. What is the probability that the number of heads is at least the number of tails?
Givens: A fair coin is tossed $4$ times, so each of the $2^4 = 16$ outcome sequences is equally likely; We want $P(\text{heads} \geq \text{tails})$ across the $4$ tosses; Answer choices: (A) $\tfrac{5}{16}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{5}{8}$, (E) $\tfrac{11}{16}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement
The phrase 'heads $\geq$ tails' on $4$ tosses splits cleanly by how many heads come up. Tool #7 (Identify Subproblems) breaks the event into three disjoint cases: exactly $2$, $3$, or $4$ heads. Tool #2 (Make a Systematic List) then counts each case by choosing which of the $4$ positions are heads. Tool #16 (Count the Complement) gives a fast check: by symmetry, the events 'more heads than tails' and 'more tails than heads' are equally likely, so the only thing to subtract from $1$ is the 'exactly tied' probability.
Execute — Answer: E
7.SP.C.7 Step 1 - Translate the condition into a cleaner one.
- With $4$ tosses, heads $+$ tails $= 4$, so 'heads $\geq$ tails' is the same as 'heads $\geq 2$', i.e.
- exactly $2$, $3$, or $4$ heads.
- These three cases do not overlap.
💡 Splitting the event into disjoint cases — $2$, $3$, or $4$ heads — is the Grade 7 'develop a uniform probability model' move.
7.SP.C.8 Step 2 - Count the total number of outcome sequences.
- Each toss is H or T, and the $4$ tosses are independent, so the sample space has $2 \times 2 \times 2 \times 2 = 16$ equally likely sequences.
💡 Listing outcomes for a compound event is the Grade 7 standard for $2$+ independent trials.
7.SP.C.8 Step 3 - Count the sequences with exactly $2$ heads.
- Pick which $2$ of the $4$ positions are heads; the other $2$ are tails.
- The systematic list (positions $1,2 \,/\, 1,3 \,/\, 1,4 \,/\, 2,3 \,/\, 2,4 \,/\, 3,4$) gives $\binom{4}{2} = 6$ sequences.
💡 Choosing $2$ positions out of $4$ to be heads is the same kind of organized list that powers Grade 7 compound-event counts.
7.SP.C.8 Step 4 - Count the sequences with exactly $3$ heads, then exactly $4$ heads.
- For $3$ heads, pick which $1$ position is the tail: $\binom{4}{1} = 4$ sequences (THHH, HTHH, HHTH, HHHT).
- For $4$ heads, only HHHH works: $\binom{4}{4} = 1$ sequence.
💡 Same listing rule applied to the other two cases — pick where the tails go.
7.SP.C.7 Step 5 - Add the three case counts and divide by the sample-space size.
- Because the cases are disjoint, the favorable count is just $6 + 4 + 1 = 11$, and each of the $16$ sequences is equally likely.
💡 Probability $=$ favorable outcomes $\div$ total equally likely outcomes — the Grade 7 uniform-model formula.
7.SP.C.7 Translate the condition into a cleaner one. With $4$ tosses, heads $+$ tails $= 7.SP.C.8 Count the total number of outcome sequences. Each toss is H or T, and the $4$ to 7.SP.C.8 Count the sequences with exactly $2$ heads. Pick which $2$ of the $4$ positions 7.SP.C.8 Count the sequences with exactly $3$ heads, then exactly $4$ heads. For $3$ head 7.SP.C.7 Add the three case counts and divide by the sample-space size. Because the cases Review
Reasonableness: Cross-check the count of $11$. There are $16$ total sequences. The sequences with strictly more tails than heads come from $0$ heads ($\binom{4}{0} = 1$, sequence TTTT) and $1$ head ($\binom{4}{1} = 4$, sequences HTTT, THTT, TTHT, TTTH), totaling $5$. So $16 - 5 = 11$ sequences have heads $\geq$ tails, giving $\tfrac{11}{16}$ — matches (E). The answer also passes a quick sanity check: half the time you'd expect a tie or a head majority, plus some bonus from ties counting in our favor, so a probability just above $\tfrac{1}{2}$ is what we should see, and $\tfrac{11}{16} = 0.6875$ fits.
Alternative: Tool #16 (Count the Complement) via symmetry. Let $p$ = P(more heads than tails), $q$ = P(more tails than heads), and $t$ = P(exactly $2$ heads). By the symmetry of a fair coin (swap H $\leftrightarrow$ T), $p = q$, and $p + q + t = 1$. From the list, $t = \tfrac{6}{16} = \tfrac{3}{8}$, so $p = q = \tfrac{1}{2}\bigl(1 - \tfrac{3}{8}\bigr) = \tfrac{5}{16}$. The asked event is 'heads $\geq$ tails' $= p + t = \tfrac{5}{16} + \tfrac{6}{16} = \tfrac{11}{16}$. Same answer (E).
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events; compare probabilities from a model to observed frequencies (Treating each of the $16$ length-$4$ H/T sequences as equally likely, so probability $=$ favorable count $\div 16$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Organizing the $16$ sequences by number of heads and counting $\binom{4}{2} = 6$, $\binom{4}{3} = 4$, $\binom{4}{4} = 1$ to get $11$ favorable outcomes.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Expressing the probability as the ratio $\tfrac{11}{16}$ of favorable outcomes to total outcomes.)
⭐ Sort the $16$ outcome sequences by how many heads showed up. The 'heads at least tails' event is $\binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 11$ out of $16$ — a Grade 7 organized-list count.
⭐ Sort the $16$ outcome sequences by how many heads showed up. The 'heads at least tails' event is $\binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 11$ out of $16$ — a Grade 7 organized-list count.
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