2018 AMC 10B Problem 6

P1-10~2 minPrint

Basic Probability Organized listing / direct enumeration

A box contains $5$ chips, numbered $1, 2, 3, 4,$ and $5$ . Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4$ . What is the probability that $3$ draws are required?

Show solution without answering

Key insight

Three draws are needed exactly when the first two chips add to 4 or less, which only happens for the pairs 1 and 2 or 1 and 3.

Solution

Three draws are needed when the first two chips do not yet add to more than $4$ , and the third chip then pushes the total over $4$ .

1. Which first two chips add to $4$ or less? Two different chips from $1$ to $5$ :
- $1 + 2 = 3$ (at most $4$ )
- $1 + 3 = 4$ (at most $4$ )
- every other pair, like $1 + 4 = 5$ or $2 + 3 = 5$ , is already more than $4$ .

So only the pairs $\{1, 2\}$ and $\{1, 3\}$ keep the drawing going.
2. Does the third draw always stop it? After $\{1,2\}$ the total is $3$ and the smallest chip left is $3$ , giving at least $6$ . After $\{1,3\}$ the total is $4$ and the smallest chip left is $2$ , giving at least $6$ . So yes, the third draw always finishes.
3. Count. The first two chips form one of $\binom{5}{2} = 10$ pairs, all equally likely. Two of them work, so the probability is
$\frac{2}{10} = \frac{1}{5}.$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{5}}$ .

Why this works

"Exactly three draws" means "not finished after two, finished after three." Here the second part always happens, so the whole question becomes counting pairs with a small sum. Just be sure the favorable cases and the total are counted the same way (both as pairs, or both in order).

Alternative approach

Count in order. The first two draws can happen in $5 \times 4 = 20$ orders. The good ones are $(1,2)$ , $(2,1)$ , $(1,3)$ , $(3,1)$ , which is $4$ orders. The probability is $\frac{4}{20} = \frac{1}{5}$ .

Concise solution

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

Three draws are required exactly when the first two draws do not already exceed $4$ , but the third draw does.

Two distinct chips from $\{1, 2, 3, 4, 5\}$ sum to at most $4$ only for the pairs $\{1, 2\}$ (sum $3$ ) and $\{1, 3\}$ (sum $4$ ). In either case the smallest remaining chip is at least $2$ , so the third draw pushes the total past $4$ automatically. Hence "three draws" is the same event as "the first two chips are $\{1,2\}$ or $\{1,3\}$ ."

There are $\binom{5}{2} = 10$ equally likely unordered pairs for the first two chips, and $2$ of them work:
$\frac{2}{10} = \frac{1}{5}.$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{5}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques