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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

adition

Cube

Check the Answer


Answer: \(\frac{1}{8}\)

AMC-10A (2003) Problem 12

Pre College Mathematics

Try with Hints


Probability in Coordinates

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……..

Coordinate Geometry

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 \(\frac{\frac{1}{2}}{4}\)=\(\frac{1}{8}\)

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