2002 AMC 10A Problem 24

P21-25~5 minPrint

Basic Probability Basic Counting Organized listing / direct enumeration Split into exhaustive cases and handle each

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

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Key insight

Tina has only 10 possible pairs. List each pair's sum; Sergio wins with every number above that sum, which is 10 minus the sum.

Solution

Tina's choices are few, so list them all. For each one, count how many of Sergio's numbers win.

1. Count all outcomes. Tina's pairs from $\{1, 2, 3, 4, 5\}$ are
$12,\ 13,\ 14,\ 15,\ 23,\ 24,\ 25,\ 34,\ 35,\ 45$
(read $12$ as "1 and 2"): $10$ pairs, all equally likely. Sergio has $10$ equally likely numbers. So there are $10 \times 10 = 100$ equally likely outcomes.

Winning numbers for one sum. If Tina's sum is $7$ , Sergio needs a number bigger than $7$ : $8, 9, 10$ . That is $3 = 10 - 7$ numbers. In general, a sum of $s$ leaves $10 - s$ winning numbers.

Go through all ten pairs.

Tina's pair	$1,2$	$1,3$	$1,4$	$1,5$	$2,3$	$2,4$	$2,5$	$3,4$	$3,5$	$4,5$
sum	$3$	$4$	$5$	$6$	$5$	$6$	$7$	$7$	$8$	$9$
Sergio's winning numbers	$7$	$6$	$5$	$4$	$5$	$4$	$3$	$3$	$2$	$1$

4. Add the bottom row:
$7 + 6 + 5 + 4 + 5 + 4 + 3 + 3 + 2 + 1 = 40.$

5. Probability:
$\frac{40}{100} = \frac{2}{5}.$

The answer is $\boxed{\textbf{(A)}\ 2/5}$ .

Why this works

When one person has only a few possible choices, fix each choice in turn and count the other person's good options. A table keeps every case in view so nothing is missed or counted twice. Pay attention to words like "larger than", which leave out ties.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Tina has $\binom{5}{2} = 10$ equally likely pairs and Sergio has $10$ equally likely numbers, so there are $100$ equally likely outcomes.

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

sum $s$	3	4	5	6	7	8	9
pairs	1	1	2	2	2	1	1
Sergio wins	7	6	5	4	3	2	1

Favorable outcomes: $1\cdot7 + 1\cdot6 + 2\cdot5 + 2\cdot4 + 2\cdot3 + 1\cdot2 + 1\cdot1 = 40$ .

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)