2009 AMC 10A Problem 22

P21-25~4 minPrint

Basic Probability Basic Counting Exploit symmetry to reduce work or pair up objects

Two cubical dice each have removable numbers $1$ through $6$ . The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is $7$ ?

Show solution without answering

Key insight

Random stickers make the first top number any of the 12, and the second any of the other 11. Exactly 2 of those 11 make 7.

Solution

The bag holds two of each number: $1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6$ .

The numbers go onto the $12$ faces in a random order, and then each die lands on a random face. So the number on top of the first die is equally likely to be any of the $12$ numbers from the bag.
That number is now used up. The top of the second die is equally likely to be any of the other $11$ numbers.
Say the first die shows $2$ . We need a $5$ on the second die. Both $5$ s are still among the $11$ numbers left (the $2$ we used is not a $5$ ). So there are $2$ good choices out of $11$ .
The same thing happens for every first number: to reach $7$ the second number must be different from the first (because $7$ is odd, it cannot be two equal numbers), so both copies of the needed number are still there. The chance is $\frac{2}{11}$ every time.

The answer is $\boxed{\textbf{(D)}\ \frac{2}{11}}$ .

Why this works

Sticking the numbers on randomly and then rolling is the same as pulling two numbers out of the bag without putting the first one back. The cubes do not matter at all. When a random setup is followed by a random choice, look for a simpler way to describe what is really being picked.

Alternative approach

Count pairs. Choose $2$ of the $12$ numbers (treat the two copies of each value as different): there are $\frac{12 \times 11}{2} = 66$ equally likely pairs. A pair adds to $7$ if it is $1$ and $6$ , $2$ and $5$ , or $3$ and $4$ . Each of these can be made in $2 \times 2 = 4$ ways, since each value has two copies. That is $12$ good pairs, and $\frac{12}{66} = \frac{2}{11}$ .

Concise solution

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

The labels are attached to the twelve faces uniformly at random, and the roll then picks one face of each die. By symmetry, the label on the first die's top face is equally likely to be any of the $12$ labels in the bag, and the label on the second die's top face is equally likely to be any of the remaining $11$ . So the pair of top labels is a uniformly random ordered pair of two distinct labels from the multiset $\{1,1,2,2,3,3,4,4,5,5,6,6\}$ .

Condition on the first label, $x$ . The sum is $7$ exactly when the second label is $7 - x$ . Because $7$ is odd, $7 - x \ne x$ , so removing $x$ leaves both copies of $7 - x$ in play: $2$ favorable labels out of $11$ .

This is the same for every $x$ , so the probability is $\dfrac{2}{11}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{2}{11}}$ .

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 Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques