AMC 8 · 2013 · #14

Easy mode Grade 7

Problem

Abe is holding $2$ jelly beans in his hand: $1$ green and $1$ red.

Bob is holding $4$ jelly beans in his hand: $1$ green, $1$ yellow, and $2$ red.

At the same moment, each of them reaches in and picks out one jelly bean at random to show the other person. Every jelly bean in a hand is equally likely to be picked.

What is the probability that the two jelly beans they show are the same color?

Pick an answer.

(A)

rac14

(B)

rac13

(C)

rac38

(D)

rac12

(E)

rac23

View mode:

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

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 Identify Subproblems 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.

$$\text{matchable colors} = \{\text{green}, \text{red}\}$$

💡 Listing the sample space and ruling out impossible outcomes is exactly the Grade 7 "find probabilities of compound events" move.

#7 Identify Subproblems 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.

$$P(\text{both green}) = \tfrac{1}{2} \times \tfrac{1}{4} = \tfrac{1}{8}$$

💡 For independent events, the chance of both happening is the product of the two chances — the Grade 7 multiplication rule.

#7 Identify Subproblems 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.

$$P(\text{both red}) = \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$$

💡 Same multiplication rule, applied to the second subproblem.

#7 Identify Subproblems 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 $8$.

$$P(\text{match}) = \tfrac{1}{8} + \tfrac{1}{4} = \tfrac{1}{8} + \tfrac{2}{8} = \tfrac{3}{8} \;\Rightarrow\; \textbf{(C)}$$

💡 Adding fractions with unlike denominators by rewriting them with a common denominator is the Grade 5 fraction-addition standard.

[1] #7 7.SP.C.8 List the colors that can possibly match. A match requires the color to be in bot

[2] #7 7.SP.C.8 Compute $P(\text{both green})$. Abe picks green with probability $\tfrac{1}{2}$

[3] #7 7.SP.C.8 Compute $P(\text{both red})$. Abe picks red with probability $\tfrac{1}{2}$ (1 o

[4] #7 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.1 Add 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.8 Find 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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