AMC 10 · 2005 · #15

Grade 7 probability
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
probability-basiccombinations-basic caseworkcomplementary-counting ↑ Prerequisites: probability-basiccombinations-basic
📏 Medium solution 💡 2 insights
Problem
An envelope holds eight bills: two 1-dollar, two 5-dollar, two 10-dollar, and two 20-dollar bills. You draw two of them at once, without replacement. Find the probability that the two bills you draw add up to 20 dollars or more.

Pick an answer.

(A)
${{{\frac{1}{4}}}}$
(B)
${{{\frac{2}{5}}}}$
(C)
${{{\frac{3}{7}}}}$
(D)
${{{\frac{1}{2}}}}$
(E)
${{{\frac{2}{3}}}}$

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Every outcome (a pair of bills) is equally likely, so the probability is just (number of good pairs) / (total pairs). Tool #7 (Identify Subproblems) splits the good pairs into two clean cases: pairs that include a 20, and pairs that do not. The first case is easiest counted with Tool #16 (Count the Complement) — count all pairs, then remove the pairs with no20. The second case is a short, finite check handled by Tool #2 (Make a Systematic List): among the remaining bills, which pairs still reach $20?

1STEP 1

Count all possible pairs

Treat the eight bills as eight distinct objects, so the total number of pairs is C(8, 2)=(8·7)/2=28 — the denominator.

C(8, 2)=(8 · 7)/2=28
2STEP 2

Any pair with a $20 already wins

A 20-dollar bill plus even the smallest bill makes 21, so every pair holding a 20 wins: 28-C(6, 2)=28-15=13 of them.

28-C(6, 2)=28-15=13
3STEP 3

Check the pairs with no $20

Among the 15 pairs with no 20, only 10+10=20 reaches the target — 10+5, 10+1, 5+5, 5+1, 1+1 fall short — so just 1 pair.

10+10=20 ✓; 10+5, 10+1, 5+5, 5+1, 1+1 ＜ 20
4STEP 4

Add the cases and form the probability

Good pairs: 13 holding a 20, plus the two-tens pair, so 13+1=14 out of 28 and P=14/28=1/2, choice (D).

P=(13+1)/28=14/28=1/2 → (D)
Answer
{{{1/2}}}
The answer 1/2 is believable: a 20 appears in a healthy share of pairs and guarantees a win, so a fifty-fifty chance is in the right ballpark — not tiny, not near-certain. A cross-check by the complement confirms it. The pairs that fall short of 20 dollars are exactly the pairs with no 20 except the two-tens pair: 15-1=14 failing pairs, matching the 28-14=14 successes. Since successes and failures are both 14, the probability is 1/2 either way. The trap choice (C) 3/7=12/28 comes from forgetting to count the 10+10 pair (or a twenty-with-twenty pair); losing two good pairs would drop 14 to 12.
💡Key takeaway

When every pick is equally likely, count the winning pairs over all pairs — and split the count into easy cases, like 'has a twenty' versus 'no twenty', so nothing gets missed or double-counted.

  • Count all possible pairs
  • Any pair with a $20 already wins
  • Check the pairs with no $20
  • Add the cases and form the probability

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