AMC 8 · 2002 · #12

Grade 7 probability

Archetype Arithmetic Expression Decomposition

probability-basicfraction-arithmeticcomplementary-counting complementary-countingidentify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 1 insight

Problem

A board game spinner is divided into three regions labeled $A$ , $B$ and $C$ . The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$ . The probability of the arrow stopping on region $C$ is:

Pick an answer.

(A)

$\frac{1}{12}$

(B)

$\frac{1}{6}$

(C)

$\frac{1}{5}$

(D)

$\frac{1}{3}$

(E)

$\frac{2}{5}$

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Toolkit + CCSS Solution

Understand

Restated: A spinner is split into three regions $A$, $B$, $C$. The arrow lands on $A$ with probability $\tfrac{1}{3}$ and on $B$ with probability $\tfrac{1}{2}$. Find the probability that it lands on $C$.

Givens: $P(A) = \dfrac{1}{3}$; $P(B) = \dfrac{1}{2}$; Regions $A$, $B$, $C$ are the only places the arrow can stop; Answer choices: (A) $\tfrac{1}{12}$, (B) $\tfrac{1}{6}$, (C) $\tfrac{1}{5}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{2}{5}$

Unknowns: $P(C)$, the probability the arrow stops on region $C$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #7 Identify Subproblems

Region $C$ is exactly the part of the spinner that is *not* $A$ and *not* $B$. Tool #16 (Count the Complement) names this directly: instead of asking "what is $P(C)$?", ask "what is left over after $A$ and $B$ take their shares of the whole?" Since all probabilities must add to $1$, the leftover is $P(C) = 1 - [P(A) + P(B)]$. Tool #7 (Identify Subproblems) splits the arithmetic into two clean pieces: first add $P(A) + P(B)$ with a common denominator, then subtract that sum from $1$.

Execute — Answer: B

#7 Identify Subproblems 5.NF.A.1 Step 1

Subproblem 1: add the two known probabilities.
To add $\tfrac{1}{3}$ and $\tfrac{1}{2}$ use the common denominator $6$.

$$P(A) + P(B) = \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6}$$

💡 Common denominator $6$ rewrites the slices so they can be combined: $\tfrac{1}{3} = \tfrac{2}{6}$ and $\tfrac{1}{2} = \tfrac{3}{6}$. Regions $A$ and $B$ together take $\tfrac{5}{6}$ of the spinner.

#16 Change Focus / Count the Complement 7.SP.C.5 Step 2

Subproblem 2: use the complement.
The whole spinner has probability $1$, and region $C$ is what is left after $A$ and $B$, so subtract their combined share from $1$.

$$P(C) = 1 - \dfrac{5}{6} = \dfrac{6}{6} - \dfrac{5}{6} = \dfrac{1}{6} \;\Rightarrow\; \textbf{(B)}$$

💡 All three probabilities add to $1$, so $P(C)$ is the missing piece. Writing $1 = \tfrac{6}{6}$ makes the subtraction work in sixths, leaving $\tfrac{1}{6}$.

[1] #7 5.NF.A.1 Subproblem 1: add the two known probabilities. To add $\tfrac{1}{3}$ and $\tfrac

[2] #16 7.SP.C.5 Subproblem 2: use the complement. The whole spinner has probability $1$, and reg

Review

Reasonableness: Check the sum: $\tfrac{1}{3} + \tfrac{1}{2} + \tfrac{1}{6} = \tfrac{2}{6} + \tfrac{3}{6} + \tfrac{1}{6} = \tfrac{6}{6} = 1$. The three probabilities add to $1$, as required for the only possible outcomes of the spin. Size check: region $C$ should be the smallest region because $A$ ($\tfrac{1}{3}$) and $B$ ($\tfrac{1}{2}$) already take most of the spinner. Indeed $\tfrac{1}{6}$ is smaller than both $\tfrac{1}{3}$ and $\tfrac{1}{2}$, which matches.

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: for each choice $P(C)$, test whether $\tfrac{1}{3} + \tfrac{1}{2} + P(C) = 1$. Only (B) $\tfrac{1}{6}$ works, since $\tfrac{2}{6} + \tfrac{3}{6} + \tfrac{1}{6} = 1$. The other four choices give totals that miss $1$, so (B) is forced.

CCSS standards used (min grade 7)

5.NF.A.1 Add and subtract fractions with unlike denominators by using equivalent fractions with a common denominator (Rewriting $\tfrac{1}{3} = \tfrac{2}{6}$ and $\tfrac{1}{2} = \tfrac{3}{6}$ to combine $P(A) + P(B) = \tfrac{5}{6}$, then computing $1 - \tfrac{5}{6} = \tfrac{1}{6}$.)
7.SP.C.5 Understand that the probability of a chance event is a number between 0 and 1, and that the probabilities of all outcomes of a chance experiment sum to 1 (Using $P(A) + P(B) + P(C) = 1$ to find $P(C)$ as the leftover probability after the two known regions.)

⭐ The whole spinner has probability $1$, and regions $A$ and $B$ already take $\tfrac{1}{3} + \tfrac{1}{2} = \tfrac{5}{6}$. Region $C$ is just the leftover: $1 - \tfrac{5}{6} = \tfrac{1}{6}$, answer (B).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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