AMC 8 · 2016 · #13

Grade 7 probabilitycounting

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccombinations-basicsystematic-enumeration systematic-enumerationcomplementary-counting ↑ Prerequisites: combinations-basicprobability-basic

📏 Short solution 💡 2 insights

Problem

Two different numbers are randomly selected from the set $\{ - 2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is $0$ ?

Pick an answer.

(A)

$dfrac{1}{6}$

(B)

$dfrac{1}{5}$

(C)

$dfrac{1}{4}$

(D)

$dfrac{1}{3}$

(E)

$dfrac{1}{2}$

AMC 8 2016 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Pick two different numbers at random from $\{-2, -1, 0, 3, 4, 5\}$ and multiply them. What is the probability that the product equals $0$?

Givens: The set has $6$ distinct numbers: $\{-2, -1, 0, 3, 4, 5\}$; Two different numbers are chosen at random; The chosen numbers are multiplied; Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{1}{5}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{2}$

Unknowns: The probability that the product of the two chosen numbers is $0$

Understand

Restated: Pick two different numbers at random from $\{-2, -1, 0, 3, 4, 5\}$ and multiply them. What is the probability that the product equals $0$?

Plan

Primary tool: #2 Make a Systematic List

Secondary: #16 Change Focus / Count the Complement

The set is small ($6$ numbers), so Tool #2 (Systematic List) can enumerate every unordered pair without any formula — easy to count favorable pairs and total pairs by hand. The favorable event collapses to a clean condition: "the pair contains $0$," because $0$ times anything is $0$ and none of the other numbers multiply to $0$. Tool #16 (Complement) gives a one-line sanity check: $P(\text{product} = 0) = 1 - P(0\text{ is NOT chosen})$, so we can confirm the answer from the opposite direction.

Execute — Answer: D

#2 Make a Systematic List 7.SP.C.8 Step 1

List every unordered pair of two different numbers from the set.
Use a fixed order — pair each number only with numbers that come after it — so nothing is double-counted or missed.

$\{-2,-1\}, \{-2,0\}, \{-2,3\}, \{-2,4\}, \{-2,5\},$ $\{-1,0\}, \{-1,3\}, \{-1,4\}, \{-1,5\},$ $\{0,3\}, \{0,4\}, \{0,5\},$ $\{3,4\}, \{3,5\}, \{4,5\}$

💡 Listing pairs in a fixed order (smaller number first, then the partner from later in the set) is the Grade 7 "organized list" move for finding sample spaces.

#2 Make a Systematic List 7.SP.C.8 Step 2

Count the total number of pairs in the list — this is the size of the sample space.

$$5 + 4 + 3 + 2 + 1 = 15 \text{ pairs}$$

💡 Counting the list directly is faster than recalling $\binom{6}{2}$ at this age, and it matches the enumeration on the page.

#2 Make a Systematic List 3.OA.B.5 Step 3

Mark the favorable pairs — those whose product is $0$.
A product is $0$ exactly when at least one factor is $0$, and only one number in the set is $0$, so every favorable pair must contain $0$.

$$\{-2,0\}, \{-1,0\}, \{0,3\}, \{0,4\}, \{0,5\} \;\Rightarrow\; 5 \text{ pairs}$$

💡 The "zero property of multiplication" ($0 \times n = 0$) is a Grade 3 property — applying it filters the list to just the pairs that include $0$.

#2 Make a Systematic List 7.SP.C.7 Step 4

Compute the probability as the ratio of favorable pairs to total pairs, then reduce.

$$P(\text{product} = 0) = \dfrac{5}{15} = \dfrac{1}{3} \;\Rightarrow\; \textbf{(D)}$$

💡 Probability $=$ (favorable outcomes) $/$ (total outcomes) when all pairs are equally likely — the Grade 7 uniform-probability model.

[1] #2 7.SP.C.8 List every unordered pair of two different numbers from the set. Use a fixed ord

[2] #2 7.SP.C.8 Count the total number of pairs in the list — this is the size of the sample spa

[3] #2 3.OA.B.5 Mark the favorable pairs — those whose product is $0$. A product is $0$ exactly

[4] #2 7.SP.C.7 Compute the probability as the ratio of favorable pairs to total pairs, then red

Review

Reasonableness: There are $6$ numbers and only $1$ of them is $0$. Picking $2$ numbers out of $6$, the chance that the special number $0$ is in the pair is $\tfrac{2}{6} = \tfrac{1}{3}$ — exactly choice (D). The size of the answer is sensible: $0$ is one of six numbers, so it should show up about a third of the time when two are picked.

Alternative: Tool #16 (Complement): $P(\text{product}=0) = 1 - P(\text{both numbers are non-zero})$. There are $\binom{5}{2} = 10$ pairs of non-zero numbers out of $15$ total, so $P(\text{both non-zero}) = \tfrac{10}{15} = \tfrac{2}{3}$, giving $1 - \tfrac{2}{3} = \tfrac{1}{3}$. Same answer.

CCSS standards used (min grade 7)

3.OA.B.5 Apply properties of operations as strategies to multiply and divide (Using the zero property of multiplication ($0 \times n = 0$) to recognize that the product of the pair is $0$ exactly when the pair contains $0$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Reducing the probability fraction $\tfrac{5}{15}$ to $\tfrac{1}{3}$ as an equivalent ratio.)
7.SP.C.7 Develop a probability model and use it to find probabilities of events (Treating each unordered pair as equally likely and computing the probability as (favorable pairs) $/$ (total pairs).)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and tree diagrams (Listing all $15$ unordered pairs in an organized sequence and counting the $5$ favorable pairs containing $0$.)

⭐ List the pairs once, find the ones with $0$, and divide — Grade 7 probability is enough to crack this AMC 8 problem.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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