Question

Point $(x,y)$ is randomly picked from the rectangular region with vertices at $(0,0),(2009,0),(2009,2010),$ and $(0,2010)$. What is the probability that $x > 7y$? Express your answer as a common fraction.

Answer 1

Hence, the probability is:

`(287)/(4020)`

Explanation:

We know that probability of an event is defined as:

Probability=(Number of favourable events)/(Total number of events)

The probability that x > 7y is given by:

Here the number of favourable event is equal to the area covered by the triangle ΔABE

Area of triangle ABE= (1/2)×b×h=(1/2)×2009×287=288291.5

and the total outcome is equal to the area of the rectangle (i.e. rectangle ABCD)

Area of rectangle ABCD=2009×2010=4038090

Hence, probability=

`(288291.5)/(4038090)=(287)/(4020)`

Answer 2

Answer: Probability that
`x>7y` is
`(287)/(4020)`

Explanation:

Since we have given that

`x>7y`

And the coordinates are as follows:

(0,0),(2009,0),(2009,2010), and (0,2010)

We need to find the probability that
`x>7y`

So, Required Probability is given by

`\frac{\text{Area of triangle}}{\text{ Area of rectangle}}\\\\=(0.5* 2009* 287)/(2009* 2010)\\\\=(287)/(4020)`

Hence, Probability that
`x>7y` is
`(287)/(4020)`