Competition · AMC preparation · step 4 of 4

AMC 8 · 2002 · #21

Grade 7 probabilitycounting

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

probability-basiccombinations-basiccomplementary-countingfraction-arithmetic caseworkcomplementary-countingsystematic-enumeration ↑ Prerequisites: combinations-basicfraction-arithmetic

📏 Short solution 💡 3 insights

Problem

Harold tosses a fair coin 4 times. What is the probability that the number of heads is at least the number of tails?

Pick an answer.

(A)

$\frac{5}{16}$

(B)

$\frac{3}{8}$

(C)

$\frac{1}{2}$

(D)

$\frac{5}{8}$

(E)

$\frac{11}{16}$

AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The phrase 'heads ≥ 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.

1STEP 1

Rewrite the condition

Since heads + tails = 4, 'heads ≥ tails' becomes 'heads ≥ 2' — exactly 2, 3, or 4 heads, three non-overlapping cases.

heads ≥ tails ⟺ heads ∈ {2, 3, 4}

2STEP 2

Count all outcomes

Each of the 4 independent tosses is H or T, so the sample space holds 2×2×2×2 = 16 equally likely sequences.

|sample space| = 2⁴ = 16

3STEP 3

Count exactly 2 heads

For exactly 2 heads, choose which 2 of the 4 positions are heads (1,2 / 1,3 / 1,4 / 2,3 / 2,4 / 3,4): C(4,2) = 6 sequences.

C(4, 2) = 6

4STEP 4

Count 3 and 4 heads

For 3 heads, pick the single tail position: C(4,3) = 4 (THHH, HTHH, HHTH, HHHT). For 4 heads, only HHHH: C(4,4) = 1.

C(4, 3) = 4, C(4, 4) = 1

5STEP 5

Add and divide

The disjoint cases give 6 + 4 + 1 = 11 favorable outcomes out of 16 equally likely sequences.

P(heads ≥ tails) = (6 + 4 + 1)/16 = 11/16 → (E)

Answer

11/16

Cross-check the count of 11. There are 16 total sequences. The sequences with strictly more tails than heads come from 0 heads (C(4, 0) = 1, sequence TTTT) and 1 head (C(4, 1) = 4, sequences HTTT, THTT, TTHT, TTTH), totaling 5. So 16 - 5 = 11 sequences have heads ≥ tails, giving

\frac{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

\frac{1}{2}

is what we should see, and

\frac{11}{16}

= 0.6875 fits.

💡Key takeaway

Sort the 16 outcome sequences by how many heads showed up. The 'heads at least tails' event is C(4, 2) + C(4, 3) + C(4, 4) = 11 out of 16 — a Grade 7 organized-list count.

Rewrite the condition
Count all outcomes
Count exactly 2 heads
Count 3 and 4 heads
Add and divide

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Rewrite the condition

Count all outcomes

Count exactly 2 heads

Count 3 and 4 heads

Add and divide

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