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Point (x, y) is randomly picked from the rectangular region with vertices at (0,0),(2008,0),(2008,2009), and (0,2009). What is the probability that x>2y?

 Feb 5, 2023
 #1
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We should draw a line from (0,2009/2) to (2008,0). Anything beneath this line is x>2y. It is a traingle so its area is  It should be put over total outcomes equal to 2008 x 2009 = 4,034,072. So we have a probability of 1,008,518\4,034,072 = 1/4.

 Feb 5, 2023
 #2
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If we try one of those points, lets say (2, 500) 2 is definatly not more than double 500.

Guest Feb 5, 2023
 #3
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To find the probability that x>2y, we need to find the area of the region where x>2y in the given rectangular region and divide it by the total area of the rectangular region.

 

The equation of the line x=2y is a line with a slope of 1/2 and y-intercept 0. The line cuts the x-axis at (0,0) and the y-axis at (2008,1004). Hence, the points of intersection with the given rectangular region are (0,0) and (2008,1004). The region where x>2y is a triangle with vertices at (0,0), (2008,0) and (2008,1004).

 

The area of the triangle can be calculated as (1/2)baseheight = (1/2)20081004 = 1004*1004.

 

The total area of the rectangular region is 2008 * 2009 = 4022416.

 

The probability that x>2y is the ratio of the area of the triangle to the total area of the rectangular region.

Hence, the required probability is:

1004 * 1004 / 4022416 = 1/4022416.

 Feb 5, 2023
 #4
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1/4 is still incorrect

Guest Feb 5, 2023
 #5
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Thank you even so

Guest Feb 5, 2023
 #6
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The stratagy is a good idea but the exucution is incorrect

Guest Feb 5, 2023

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