AMC 8 · 2013 · #14

Grade 7 probabilitycounting

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsystematic-enumerationfraction-arithmetic caseworksystematic-enumeration ↑ Prerequisites: fraction-arithmeticsystematic-enumeration

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Problem

Abe is holding 2 jelly beans (1 green, 1 red). Bob is holding 4 jelly beans (1 green, 1 yellow, 2 red). Each one picks a jelly bean uniformly at random and shows it to the other. What is the probability that the two shown colors are the same?

Pick an answer.

(A)

$\frac14$

(B)

$\frac13$

(C)

$\frac38$

(D)

$\frac12$

(E)

$\frac23$

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

A match can happen in two completely separate ways — both pick green, or both pick red — so Tool #7 (Identify Subproblems) lets us solve each case on its own and add the results, since the two cases cannot happen at the same time. Tool #4 (Make a Systematic List) gives a useful cross-check: there are 2 × 4 = 8 equally likely (Abe-pick, Bob-pick) pairs, and we can just count how many of those 8 pairs have matching colors.

1STEP 1

A match needs the color in both hands. Abe has {green, red}, Bob {green, yellow, red}, so only green and red can match — two cases.

matchable colors = {green, red}

2STEP 2

Abe picks green with chance $\frac{1}{2}$ , Bob with $\frac{1}{4}$ . Picks are independent, so multiply: P(both green) = $\frac{1}{8}$ .

P(both green) =

\frac{1}{2}

\frac{1}{4}

\frac{1}{8}

3STEP 3

Abe picks red with chance $\frac{1}{2}$ , Bob with $\frac{2}{4}$ = $\frac{1}{2}$ (two red beans). Multiply: P(both red) = $\frac{1}{4}$ .

P(both red) =

\frac{1}{2}

\frac{1}{2}

\frac{1}{4}

4STEP 4

The two cases can't both happen, so add them: $\frac{1}{8}$ + $\frac{1}{4}$ = $\frac{1}{8}$ + $\frac{2}{8}$ = $\frac{3}{8}$ , choice (C).

P(match) =

\frac{1}{8}

\frac{1}{4}

\frac{1}{8}

\frac{2}{8}

\frac{3}{8}

→ (C)

Answer

\frac{3}{8}

Yellow is dead weight: Bob has a

\frac{1}{4}

chance of picking yellow, in which case no match is possible. So the match probability must be strictly less than

\frac{3}{4}

, and intuitively well under

\frac{1}{2}

since Abe only has 2 colors to begin with. Our answer

\frac{3}{8}

= 0.375 sits naturally between

\frac{1}{4}

and

\frac{1}{2}

, matching choice (C).

💡Key takeaway

This AMC 8 problem only needs the Grade 7 idea that you multiply chances for independent picks and add chances for cases that can't both happen!

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