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

Answer: Probability that $x>7y$ is $\frac{287}{4020}$

Step-by-step 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

$\frac{\text{Area of triangle}}{\text{ Area of rectangle}}\\\\=\frac{0.5\times 2009\times 287}{2009\times 2010}\\\\=\frac{287}{4020}$

Hence, Probability that $x>7y$ is $\frac{287}{4020}$

Answer 2

Answer:

Hence, the probability is:

$\dfrac{287}{4020}$

Step-by-step 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=

$\dfrac{288291.5}{4038090}=\dfrac{287}{4020}$