AMC 10 · 2011 · #13
Grade 7 probabilityTwo real numbers are selected independently at random from the interval [−20,10]. What is the probability that the product of those numbers is greater than zero?
Pick an answer.
AMC 10 2011 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: Two numbers are picked at random, each anywhere in the interval $[-20, 10]$, and the picks do not affect each other. Find how likely it is that the two numbers multiply to something greater than zero.
Givens: Each number is chosen uniformly at random from $[-20, 10]$; The two choices are independent of each other; Answer choices: (A) $\frac{1}{9}$, (B) $\frac{1}{3}$, (C) $\frac{4}{9}$, (D) $\frac{5}{9}$, (E) $\frac{2}{3}$
Unknowns: The probability that the product of the two numbers is greater than zero
Understand
Restated: Two numbers are picked at random, each anywhere in the interval $[-20, 10]$, and the picks do not affect each other. Find how likely it is that the two numbers multiply to something greater than zero.
Givens: Each number is chosen uniformly at random from $[-20, 10]$; The two choices are independent of each other; Answer choices: (A) $\frac{1}{9}$, (B) $\frac{1}{3}$, (C) $\frac{4}{9}$, (D) $\frac{5}{9}$, (E) $\frac{2}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #1 Draw a Diagram
The key reframe (Tool #16) is that “product greater than zero” just means “both numbers have the same sign,” which splits cleanly into two separate jobs (Tool #7): both numbers negative, or both numbers positive. Each job is a small independent-probability calculation, and the two jobs never overlap, so their chances add. Tool #1 (Draw a Diagram) backs this up: picturing the interval as a number line makes the negative part (length $20$) and positive part (length $10$) easy to measure against the whole length $30$.
Execute — Answer: D
7.SP.C.7 Step 1 Measure the negative and positive parts
- Lay $[-20, 10]$ on a number line.
- Its total length is $10-(-20)=30$.
- The negative part runs from $-20$ to $0$, length $20$; the positive part runs from $0$ to $10$, length $10$.
- So for one pick, the chance it is negative is $\frac{20}{30}=\frac{2}{3}$ and the chance it is positive is $\frac{10}{30}=\frac{1}{3}$.
💡 On a uniform interval, the chance of a region is just its length divided by the whole length.
7.SP.C.8 Step 2 Chance both picks are negative
- For the product to be positive one way is that both numbers are negative.
- The two picks are independent, so multiply their chances: $\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$.
💡 For independent events, the chance of both happening is the product of their separate chances.
7.SP.C.8 Step 3 Chance both picks are positive
- The other way to get a positive product is that both numbers are positive.
- Again multiply the independent chances: $\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$.
💡 Same-sign multiplied gives positive, so the both-positive case counts too.
4.NF.B.3 Step 4 Add the two separate cases
- Both-negative and both-positive can never happen at the same time, so add their chances: $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$.
- That is the probability the product is greater than zero, which is choice (D).
💡 Chances of separate, non-overlapping cases add together.
7.SP.C.7 Lay $[-20, 10]$ on a number line. Its total length is $10-(-20)=30$. The negativ 7.SP.C.8 For the product to be positive one way is that both numbers are negative. The tw 7.SP.C.8 The other way to get a positive product is that both numbers are positive. Again 4.NF.B.3 Both-negative and both-positive can never happen at the same time, so add their Review
Reasonableness: Check with the complement. The product is negative when the two picks have opposite signs, which happens two ways: neg-then-pos or pos-then-neg, giving $2\times\frac{2}{3}\times\frac{1}{3}=\frac{4}{9}$. Since the chance of landing exactly on $0$ is $0$, positive and negative products must fill the whole probability: $\frac{5}{9}+\frac{4}{9}=1$. It checks out, and $\frac{5}{9}$ is more than half, which makes sense because the fatter negative side raises the chance of a same-sign match.
Alternative: Draw the sample space as a $30\times30$ square (one pick on each axis). The positive-product region is the bottom-left $20\times20$ block (both negative, area $400$) plus the top-right $10\times10$ block (both positive, area $100$). That is $\frac{400+100}{900}=\frac{500}{900}=\frac{5}{9}$, matching (D).
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Turning region lengths into single-pick chances: $P(\text{neg})=\frac{2}{3}$ and $P(\text{pos})=\frac{1}{3}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Multiplying independent picks to get $P(\text{both neg})=\frac{4}{9}$ and $P(\text{both pos})=\frac{1}{9}$.)4.NF.B.3Add and subtract fractions with like denominators (Combining the two disjoint cases: $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$.)
⭐ A product is positive only when both numbers share a sign, so find each same-sign chance and add them up.
⭐ A product is positive only when both numbers share a sign, so find each same-sign chance and add them up.
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