AMC 8 · 2013 · #14
Grade 7 probabilitycountingProblem
Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
Pick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Abe's hand: $1$ green $+$ $1$ red ($2$ total); Bob's hand: $1$ green $+$ $1$ yellow $+$ $2$ red ($4$ total); Each picks one jelly bean uniformly at random, independently of the other; Answer choices: (A) $\tfrac14$, (B) $\tfrac13$, (C) $\tfrac38$, (D) $\tfrac12$, (E) $\tfrac23$
Unknowns: The probability that Abe's color matches Bob's color
Understand
Restated: 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?
Givens: Abe's hand: $1$ green $+$ $1$ red ($2$ total); Bob's hand: $1$ green $+$ $1$ yellow $+$ $2$ red ($4$ total); Each picks one jelly bean uniformly at random, independently of the other; Answer choices: (A) $\tfrac14$, (B) $\tfrac13$, (C) $\tfrac38$, (D) $\tfrac12$, (E) $\tfrac23$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Make a Systematic List
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 $2 \times 4 = 8$ equally likely (Abe-pick, Bob-pick) pairs, and we can just count how many of those $8$ pairs have matching colors.
Execute — Answer: C
7.SP.C.8 Step 1 - List the colors that can possibly match.
- A match requires the color to be in both hands.
- Abe has $\{\text{green}, \text{red}\}$ and Bob has $\{\text{green}, \text{yellow}, \text{red}\}$, so the shared colors are green and red.
- Yellow is impossible to match, so we only need two subproblems: both-green and both-red.
💡 Listing the sample space and ruling out impossible outcomes is exactly the Grade 7 "find probabilities of compound events" move.
7.SP.C.8 Step 2 - Compute $P(\text{both green})$.
- Abe picks green with probability $\tfrac{1}{2}$ (1 out of his 2 beans).
- Bob picks green with probability $\tfrac{1}{4}$ (1 out of his 4 beans).
- Their picks are independent, so multiply.
💡 For independent events, the chance of both happening is the product of the two chances — the Grade 7 multiplication rule.
7.SP.C.8 Step 3 - Compute $P(\text{both red})$.
- Abe picks red with probability $\tfrac{1}{2}$ (1 out of 2).
- Bob picks red with probability $\tfrac{2}{4} = \tfrac{1}{2}$ (2 out of 4).
- Multiply again.
💡 Same multiplication rule, applied to the second subproblem.
5.NF.A.1 Step 4 - Add the two subproblem probabilities.
- The events "both green" and "both red" cannot happen at the same time (one color per draw), so the chance of a match is the sum.
- Use a common denominator of $8$.
💡 Adding fractions with unlike denominators by rewriting them with a common denominator is the Grade 5 fraction-addition standard.
7.SP.C.8 List the colors that can possibly match. A match requires the color to be in bot 7.SP.C.8 Compute $P(\text{both green})$. Abe picks green with probability $\tfrac{1}{2}$ 7.SP.C.8 Compute $P(\text{both red})$. Abe picks red with probability $\tfrac{1}{2}$ (1 o 5.NF.A.1 Add the two subproblem probabilities. The events "both green" and "both red" can Review
Reasonableness: Yellow is dead weight: Bob has a $\tfrac{1}{4}$ chance of picking yellow, in which case no match is possible. So the match probability must be strictly less than $\tfrac{3}{4}$, and intuitively well under $\tfrac{1}{2}$ since Abe only has $2$ colors to begin with. Our answer $\tfrac{3}{8} = 0.375$ sits naturally between $\tfrac{1}{4}$ and $\tfrac{1}{2}$, matching choice (C).
Alternative: Tool #4 (Make a Systematic List): label Bob's beans $G, Y, R_1, R_2$. The $2 \times 4 = 8$ equally likely (Abe, Bob) pairs are $(G,G), (G,Y), (G,R_1), (G,R_2), (R,G), (R,Y), (R,R_1), (R,R_2)$. The matching pairs are $(G,G), (R,R_1), (R,R_2)$ — exactly $3$ out of $8$, giving $\tfrac{3}{8}$.
CCSS standards used (min grade 7)
5.NF.A.1Add and subtract fractions with unlike denominators (Adding $\tfrac{1}{8} + \tfrac{1}{4}$ by rewriting with a common denominator of $8$ to get $\tfrac{3}{8}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Treating Abe's and Bob's picks as independent events, computing the probability of each matching scenario as a product, and combining the two mutually exclusive cases.)
⭐ 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!
⭐ 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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