2006 AMC 10B Problem 17

P16-20~2 minPrint

Basic Probability Conditional Probability & States Exploit symmetry to reduce work or pair up objects

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?

Show solution without answering

Key insight

Whatever color Alice gives, Bob then has six balls with two of that color. The bags match only if Bob sends back one of those two.

Solution

Try a concrete case. Say Alice picks green. Now Alice has blue, orange, red, violet (no green). Bob has six balls: blue, green, green, orange, red, violet.
When can the bags end up the same? Alice is missing green, so she needs a green back. If Bob sends green, both bags again have one ball of each color.
If Bob sends any other color, say red, Alice ends with two reds and no green, while Bob still has two greens. The bags are different.
Bob picks one of his $6$ balls at random, and $2$ of them are green. So the chance is $\frac{2}{6} = \frac{1}{3}$ .
Nothing here depended on green. Whatever color Alice picks, Bob ends up with two of that color among six balls, and the chance is again $\frac13$ .

The answer is $\boxed{\textbf{(D)}\ \frac{1}{3}}$ .

Why this works

Work through one specific first move. When all the first moves look alike (here, all colors are treated the same), the answer for one case is the answer overall. Also notice that only the colors matter, not which physical ball moves, so there are $2$ good balls, not $1$ .

Concise solution

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

Suppose Alice transfers a ball of some color, say green. Alice now has four colors (no green); Bob has six balls, two of which are green.

For the bags to end up identical, Alice must get a green ball back: then both bags again hold one of each color. If Bob sends any other color, Alice ends up with two of that color and no green while Bob has two greens, so the bags differ.

Bob picks uniformly among his six balls, and two of them are green, so the probability is $\frac{2}{6} = \frac{1}{3}$ .

Because this reasoning holds for every color Alice might choose, the overall probability is also $\frac{1}{3}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{1}{3}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Intermediate Counting & Probability, Chapter 11: Conditional Probability