AMC 10 · 2003 · #12
Grade 7 geometry-2dA point (x,y) is randomly picked from inside the rectangle with vertices (0,0), (4,0), (4,1), and (0,1). What is the probability that x<y?
Pick an answer.
AMC 10 2003 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: A point $(x,y)$ is chosen at random, evenly, from inside the rectangle whose corners are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. Find the probability that the point's $x$-coordinate is smaller than its $y$-coordinate, that is, the probability that $x<y$.
Givens: The point is picked uniformly at random from inside a rectangle.; The rectangle spans $x$ from $0$ to $4$ and $y$ from $0$ to $1$.; We want the chance that the condition $x<y$ holds.; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{1}{4}$, (C) $\frac{3}{8}$, (D) $\frac{1}{2}$, (E) $\frac{3}{4}$.
Unknowns: The probability that a uniformly random point in the rectangle satisfies $x<y$.
Understand
Restated: A point $(x,y)$ is chosen at random, evenly, from inside the rectangle whose corners are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. Find the probability that the point's $x$-coordinate is smaller than its $y$-coordinate, that is, the probability that $x<y$.
Givens: The point is picked uniformly at random from inside a rectangle.; The rectangle spans $x$ from $0$ to $4$ and $y$ from $0$ to $1$.; We want the chance that the condition $x<y$ holds.; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{1}{4}$, (C) $\frac{3}{8}$, (D) $\frac{1}{2}$, (E) $\frac{3}{4}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #14 Extreme Principle
When a point is chosen evenly at random from a region, probability is just a ratio of areas: (area of the good part) over (area of the whole). So Tool #1 (Draw a Diagram) is the heart of the work — sketch the rectangle, draw the boundary line $y=x$, and shade the piece where $x<y$. Tool #7 (Identify Subproblems) splits the job into two clean pieces: find the total area, then find the shaded area. Tool #14 (Extreme Principle) does the sharp thinking up front: since $y$ never exceeds $1$, the inequality $x<y$ forces $x<1$, so the entire long right end of the rectangle (from $x=1$ to $x=4$) is dead — the good region is trapped in the first unit of width.
Execute — Answer: A
6.G.A.1 Step 1 Total area is the sample space
- Because the point is chosen uniformly, every equal patch of the rectangle is equally likely, so probability is the fraction of the rectangle's area that satisfies the condition.
- First measure the whole rectangle: it is $4$ wide (from $x=0$ to $x=4$) and $1$ tall (from $y=0$ to $y=1$), so its area is $4\times 1 = 4$.
- This $4$ is the denominator of the probability.
💡 With an even sprinkle of points, the chance of landing in a region is simply how big a slice of the total area that region takes up.
6.G.A.3 Step 2 Locate the region where x < y
- Draw the line $y=x$.
- Points with $x<y$ lie above this line.
- Inside the rectangle, the top edge is $y=1$, and the line $y=x$ reaches that top edge at $(1,1)$; it enters at the corner $(0,0)$.
- So above the line and inside the rectangle we get the triangle with corners $(0,0)$, $(0,1)$, and $(1,1)$.
- Everything to the right of $x=1$ has $x\ge 1>y$, so none of that long strip counts — the good region really is just this small triangle in the upper-left.
💡 Since heights only reach $1$, an $x$ bigger than $1$ can never be beaten by $y$, which pins the whole winning region into the leftmost unit of width.
6.G.A.1 Step 3 Area of the winning triangle
- The triangle with corners $(0,0)$, $(0,1)$, and $(1,1)$ is a right triangle: the left edge from $(0,0)$ to $(0,1)$ is vertical with length $1$, and the top edge from $(0,1)$ to $(1,1)$ is horizontal with length $1$.
- These two legs meet at a right angle at $(0,1)$, so the area is half the base times the height: $\tfrac12\times 1\times 1 = \tfrac12$.
- This $\tfrac12$ is the numerator of the probability.
💡 A right triangle is exactly half of the rectangle you could draw on its two legs, so its area is half of $1\times 1$.
7.SP.C.7 Step 4 Divide the areas to get the probability
- Probability is the good area over the total area: $\dfrac{1/2}{4} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}$.
- So a random point has a $\dfrac{1}{8}$ chance of satisfying $x<y$, which is choice (A).
💡 Turning a random-point question into a division of two areas is what makes geometric probability easy to finish.
6.G.A.1 Because the point is chosen uniformly, every equal patch of the rectangle is equ 6.G.A.3 Draw the line $y=x$. Points with $x<y$ lie above this line. Inside the rectangle 6.G.A.1 The triangle with corners $(0,0)$, $(0,1)$, and $(1,1)$ is a right triangle: the 7.SP.C.7 Probability is the good area over the total area: $\dfrac{1/2}{4} = \dfrac{1}{2} Review
Reasonableness: The winning triangle is tiny compared to the long rectangle: it fits inside the first $1\times 1$ square on the left, and even there it is only half of that square. One unit square is $\tfrac14$ of the whole rectangle, and half of that is $\tfrac18$ — a small probability, which fits the picture of a skinny wide rectangle where $x$ is usually far bigger than the at-most-$1$ value of $y$. So the small answer $\tfrac18$ is believable, and the larger choices like $\tfrac12$ or $\tfrac34$ would be far too big.
Alternative: Introduce a variable and integrate (Tool #4/#13 flavor): for the condition $x<y$ with $0\le y\le 1$, the allowed $x$ runs from $0$ to $y$, a width of $y$. Averaging that width over $y$ from $0$ to $1$ gives area $\int_0^1 y\,dy = \tfrac12$, and dividing by the total area $4$ again yields $\tfrac{1}{8}$ — the same answer without drawing the triangle.
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles and rectangles by composing or decomposing shapes (Computing the rectangle's area ($4$) and the winning right triangle's area ($\tfrac12$).)6.G.A.3Draw polygons in the coordinate plane and use coordinates to find side lengths (Plotting the line $y=x$ and the rectangle to identify the triangular region where $x<y$ and its vertices.)7.SP.C.7Develop a uniform probability model and use it to find probabilities of events (Treating the uniform random point as a ratio of areas: probability equals winning area over total area.)
⭐ When a point is dropped evenly on a region, the probability is just the good area divided by the whole area — here a $\tfrac12$ triangle inside an area-$4$ rectangle gives $\tfrac{1}{8}$.
⭐ When a point is dropped evenly on a region, the probability is just the good area divided by the whole area — here a $\tfrac12$ triangle inside an area-$4$ rectangle gives $\tfrac{1}{8}$.
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