AMC 10 · 2004 · #10

Grade 7 probability
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
probability-basiccombinations-basic caseworkidentify-subproblems ↑ Prerequisites: fraction-arithmetic
📏 Medium solution 💡 2 insights
Problem
A fair coin A is flipped 3 times and a fair coin B is flipped 4 times. Find the probability that coin A shows the same number of heads as coin B.

Pick an answer.

(A)
$\frac{19}{128}$
(B)
$\frac{23}{128}$
(C)
$\frac14$
(D)
$\frac{35}{128}$
(E)
$\frac12$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

"Same number of heads" is not one event but several rolled together: both show 0, both show 1, both show 2, or both show 3. That is a signal to break the problem into cases (Tool #7) — one subproblem per shared head-count — and add the results, since the cases cannot happen at the same time. Inside each case the coins are independent, so the count of matching flip-sequences is (ways A gets that many heads) × (ways B gets that many heads). A systematic list (Tool #2) sweeping k = 0, 1, 2, 3 makes sure no case is skipped or double-counted. Working in whole counts of equally likely sequences, then dividing by the total at the end, keeps the arithmetic clean.

1STEP 1

Count all equally likely outcomes

Coin A's 3 flips give 8 sequences and coin B's 4 give 16; independent pairing makes 128 equally likely outcomes.

2³ × 2⁴ = 8 × 16 = 128
2STEP 2

List the head-counts that can match

Coin A tops out at 3 heads, so the counts agree only at 0, 1, 2, or 3 — coin B's fourth head can never be matched.

matching counts k ∈ {0, 1, 2, 3}
3STEP 3

Count matching sequences in each case

Independence lets you multiply inside a case: C(3, k) times C(4, k) gives 1, 12, 18, and 4 matching pairs.

C(3, k)C(4, k): 1·1=1, 3·4=12, 3·6=18, 1·4=4
4STEP 4

Add the cases and form the probability

The four cases cannot overlap, so add: 1 + 12 + 18 + 4 = 35 matches out of 128, that is 35128\frac{35}{128} — choice (D).

(1+12+18+4)/128 = 35/128 → (D)
Answer
35/128
The value 35/128 ≈ 0.27 is a little above 1/4, which fits the picture: matching is fairly common but far from guaranteed, so an answer near a quarter is believable while 1/2 (E) would be far too high. The choices also expose the traps — rounding to the clean-looking 1/4 gives (C), and slips in the case counts land on the near-miss numerators 19 and 23 over 128. Only the exact count 1+12+18+4 = 35 produces (D).
💡Key takeaway

When two coin batches must match, handle one shared head-count at a time: multiply each batch's ways within a case, add the cases, and put it over all the equally likely outcomes.

  • Count all equally likely outcomes
  • List the head-counts that can match
  • Count matching sequences in each case
  • Add the cases and form the probability

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