AMC 10 · 2002 · #16

Grade 7 probability
Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split
probability-basiccombinations-basicsystematic-enumeration casework ↑ Prerequisites: probability-basic
📏 Long solution 💡 3 insights
Problem
Tina picks two different numbers from {1,2,3,4,5}, with every pair equally likely. Independently, Sergio picks one number from {1,2,…,10}, with every number equally likely. Find the probability that Sergio's number is strictly larger than the sum of Tina's two numbers.

Pick an answer.

(A)
2/5
(B)
9/20
(C)
1/2
(D)
11/20
(E)
24/25
How to solve
Strategy Make a Systematic List

The whole experiment has only 100 equally likely outcomes, so Tool #2 (Make a Systematic List) can reach the answer by honest counting — no clever trick required. The list is kept short by Tool #7 (Identify Subproblems): all that matters about Tina's pair is its sum, so pairs are grouped by sum and each group handled once. Tool #4 (Introduce a Variable) supplies the bridge — call Tina's sum T and Sergio's winning count is 10-T every single time, which turns seven separate cases into one formula.

1STEP 1

Count all equally likely outcomes

Tina has C(5,2) = 10 pairs and Sergio 10 numbers, so there are 100 equally likely outcomes.

C(5, 2) × 10 = 10 × 10 = 100
2STEP 2

How many numbers beat a given sum

If Tina's sum is T, exactly 10 - T of Sergio's numbers beat it, and T runs from 3 to 9.

#{s : T ＜ s ≤ 10} = 10 - T, 3 ≤ T ≤ 9
3STEP 3

Group Tina's pairs by their sum

Sorting the pairs by sum gives group sizes 1,1,2,2,2,1,1, which add to all 10 pairs.

3{:} 1, 4{:} 1, 5{:} 2, 6{:} 2, 7{:} 2, 8{:} 1, 9{:} 1 (1+1+2+2+2+1+1 = 10)
4STEP 4

Add up the favourable outcomes

Weighting each sum by its pair count gives 7+6+10+8+6+2+1 = 40 favourable outcomes.

1(7) + 1(6) + 2(5) + 2(4) + 2(3) + 1(2) + 1(1) = 40
5STEP 5

Divide and reduce

So the probability is 40/100 = 2/5, choice (A).

40/100 = 2/5 → (A)
Answer
2/5
Tina's sum sits in the middle of Sergio's range — its smallest value is 3 and its largest is 9, centred on 6 — so Sergio should win noticeably less than half the time, since he loses on every tie and on everything below. That already pushes (C) 1/2, (D) 11/20 and (E) 24/25 out of range, leaving something near 2/5. The expected-value route below reaches 2/5 without listing a single case, so two independent methods agree on (A).
💡Key takeaway

When every outcome is equally likely, sort them into groups that behave the same way, count each group once, and the probability is favourable over total.

  • Count all equally likely outcomes
  • How many numbers beat a given sum
  • Group Tina's pairs by their sum
  • Add up the favourable outcomes
  • Divide and reduce

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