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# Probability in Coordinates | AMC-10A, 2003 | Problem 12

Try this beautiful problem from Probability in Coordinates from AMC-10A, 2003. You may use sequential hints to solve the problem.

Try this beautiful problem from Probability based on Coordinates.

## Probability in Coordinates – AMC-10A, 2003- Problem 12

A 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 $x<y$?

• $\frac{1}{8}$
• $\frac{1}{6}$
• $\frac{2}{3}$

### Key Concepts

Number system

Cube

Answer: $\frac{1}{8}$

AMC-10A (2003) Problem 12

Pre College Mathematics

## Try with Hints

The given vertices are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of $OC$= $4$ and length of $AO$=$1$.Therefore area of the rectangle is $4 \times 1=4$.now we have to find out the probability that $x<y$.so we draw a line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.can you find out the area with the condition $x<y$?

can you finish the problem……..

Now the line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.Therefore it will form a Triangle $\triangle AOD$ (as shown above) whose $AO=1$ and $AD=1$.Therefore area of $\triangle AOD=\frac{1}{2}$ i.e (red region).Now can you find out the probability with the condition $x<y$?

can you finish the problem……..

Therefore the required probability ($x<y$) is $\frac{\frac{1}{2}}{4}$=$\frac{1}{8}$