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2002 AMC 10A Problem 24

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2002 AMC 10A Problem 24AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 4 of 5 · about 5 min · position P21-25

Tina randomly selects two distinct numbers from the set {1,2,3,4,5}\{ 1, 2, 3, 4, 5 \} , and Sergio randomly selects a number from the set {1,2,...,10}\{ 1, 2, ..., 10 \} . What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

  1. A)

    2/52/5

  2. B)

    9/209/20

  3. C)

    1/21/2

  4. D)

    11/2011/20

  5. E)

    24/2524/25

Answer

A

Key insight

Tina has only 10 possible pairs; tabulate each pair's sum s, and Sergio wins in exactly 10 - s ways, so the answer is a short sum divided by 100.

Solution

Tina has (52)=10\binom{5}{2} = 10 equally likely pairs and Sergio has 1010 equally likely numbers, so there are 100100 equally likely outcomes.

If Tina's sum is ss, Sergio's number must be one of s+1,…,10s+1, \dots, 10: that is 10−s10 - s choices. Tabulate the sums:

sum ss3456789
pairs1122211
Sergio wins7654321

Favorable outcomes: 1⋅7+1⋅6+2⋅5+2⋅4+2⋅3+1⋅2+1⋅1=401\cdot7 + 1\cdot6 + 2\cdot5 + 2\cdot4 + 2\cdot3 + 1\cdot2 + 1\cdot1 = 40.

The probability is 40100=25\frac{40}{100} = \frac{2}{5}, so the answer is (A) 2/5\boxed{\textbf{(A)}\ 2/5}.

Why this works

When one player's choice is small and structured (only 1010 pairs), fix that choice and count the other player's winning options as a simple function of it. A two-row table keeps the casework honest and makes the multiplication explicit.

The trap

Using 'at least' instead of 'strictly larger' (counting 11 - s instead of 10 - s), or miscounting how many pairs give each sum.

Common mistakes

  • Using 'at least' instead of 'strictly larger' (counting 11 - s instead of 10 - s), or miscounting how many pairs give each sum.
  • Treating Tina's pairs as ordered (2020 outcomes) for the total but unordered for the favorable count.

Techniques

Organized listing / direct enumeration · Split into exhaustive cases and handle each