AMC 10 · 2004 · #5
Grade 7 probabilityA set of three points is randomly chosen from the grid shown. Each three point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?
Pick an answer.
AMC 10 2004 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: The figure is a $3\times3$ square grid of $9$ dots. Three of the dots are chosen at random, with every set of three dots equally likely. Find the probability that the three chosen dots all lie on one straight line.
Givens: The grid has $9$ dots arranged in $3$ rows and $3$ columns.; Exactly three dots are chosen, and every three-dot set is equally likely.; Answer choices: (A) $\frac{1}{21}$, (B) $\frac{1}{14}$, (C) $\frac{2}{21}$, (D) $\frac{1}{7}$, (E) $\frac{2}{7}$
Unknowns: The probability that the three chosen dots are collinear (lie on one straight line).
Understand
Restated: The figure is a $3\times3$ square grid of $9$ dots. Three of the dots are chosen at random, with every set of three dots equally likely. Find the probability that the three chosen dots all lie on one straight line.
Givens: The grid has $9$ dots arranged in $3$ rows and $3$ columns.; Exactly three dots are chosen, and every three-dot set is equally likely.; Answer choices: (A) $\frac{1}{21}$, (B) $\frac{1}{14}$, (C) $\frac{2}{21}$, (D) $\frac{1}{7}$, (E) $\frac{2}{7}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #1 Draw a Diagram
Because every three-dot set is equally likely, the probability is just (collinear sets) $\div$ (all sets), so the problem splits into two separate counts (Tool #7): a denominator (how many ways to pick $3$ dots from $9$) and a numerator (how many of those picks are straight lines). The denominator is a plain "choose $3$ from $9$" count. The numerator is where care is needed, so a systematic list (Tool #2) sweeps the grid by direction — rows, then columns, then diagonals — with a quick diagram (Tool #1) so no line is missed or double-counted.
Execute — Answer: C
7.SP.C.7 Step 1 Set up probability as a fraction
- Every three-dot set is equally likely, so the probability of a straight line is the number of three-dot sets that form a line divided by the total number of three-dot sets.
- This turns the whole problem into two counting jobs: count all the sets for the bottom of the fraction, and count the straight-line sets for the top.
💡 When every outcome is equally likely, probability is simply the fraction of outcomes you want out of all outcomes.
7.SP.C.8 Step 2 Count all three-dot sets
- To pick $3$ dots out of $9$, order does not matter.
- Counting ordered picks gives $9\times8\times7=504$, but each group of $3$ dots gets counted $3\times2\times1=6$ times (once for each order), so divide: $504\div6=84$.
- There are $84$ equally likely three-dot sets, so the bottom of the fraction is $84$.
💡 Picking a group where order does not matter means count the ordered ways, then divide by the number of orderings of each group.
7.SP.C.8 Step 3 List the straight lines through three dots
- Sweep the grid by direction so nothing is missed.
- Horizontal lines: the top, middle, and bottom rows give $3$ lines.
- Vertical lines: the left, middle, and right columns give $3$ more.
- Diagonal lines: only the two full corner-to-corner diagonals pass through $3$ dots, giving $2$.
- Any other slanted line (for example a gentle diagonal) hits at most $2$ of the dots, so it does not count.
- That is $3+3+2=8$ straight-line sets.
💡 Organizing the search by direction — rows, columns, diagonals — guarantees every line is found exactly once.
6.NS.B.4 Step 4 Divide and simplify
- The probability is the $8$ straight-line sets over the $84$ total sets: $\frac{8}{84}$.
- The greatest common factor of $8$ and $84$ is $4$, so divide top and bottom by $4$ to get $\frac{2}{21}$.
- That matches choice (C).
💡 Dividing the numerator and denominator by their greatest common factor writes the probability in lowest terms.
7.SP.C.7 Every three-dot set is equally likely, so the probability of a straight line is 7.SP.C.8 To pick $3$ dots out of $9$, order does not matter. Counting ordered picks gives 7.SP.C.8 Sweep the grid by direction so nothing is missed. Horizontal lines: the top, mid 6.NS.B.4 The probability is the $8$ straight-line sets over the $84$ total sets: $\frac{8 Review
Reasonableness: The probability $\frac{2}{21}$ is about $0.10$, a small chance — which fits the picture, since three dots dropped randomly rarely land in a perfect line. The counting also explains the trap answers: forgetting the two diagonals leaves $6$ lines and gives $\frac{6}{84}=\frac{1}{14}$, which is the tempting choice (B). Only when the diagonals are included does the count reach $8$ and the probability become $\frac{2}{21}$, confirming (C).
Alternative: Instead of the "order does not matter, then divide by $6$" shortcut for the denominator, a student can reason directly: there are $8$ known lines, and the total number of triples is the combination $\binom{9}{3}=84$. Reaching $\frac{8}{84}=\frac{2}{21}$ by this route confirms (C) and shows the answer does not depend on the counting trick used.
CCSS standards used (min grade 7)
7.SP.C.7Develop a uniform probability model and use it to find probabilities of events (Turning the equally-likely setup into the fraction (collinear sets) / (all sets).)7.SP.C.8Find probabilities of compound events using organized lists, tables, and counting (Counting the 84 total three-dot sets and systematically listing the 8 straight-line sets.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Reducing the probability $\frac{8}{84}$ to lowest terms $\frac{2}{21}$ using the GCF $4$.)
⭐ For a probability with equally likely picks, count the winners over the total — and count straight lines by sweeping rows, then columns, then diagonals so you never miss one.
⭐ For a probability with equally likely picks, count the winners over the total — and count straight lines by sweeping rows, then columns, then diagonals so you never miss one.
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