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In the x- y plane, there are 4 points (0,0), (0,4), (6,4), a

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by Max@Math Revolution » Mon Jun 13, 2016 9:02 am
In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). If these 4 points makes a rectangle, what is the probability that x+y<4?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 2/5

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by 800_or_bust » Mon Jun 13, 2016 9:21 am
Max@Math Revolution wrote:In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). If these 4 points makes a rectangle, what is the probability that x+y<4?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 2/5

*Answer will be posted in 2 days.
I'm assuming x and y are the x- and y-coordinates of an ordered pair (x,y) lying inside the rectangle and selected at random.

The total area of the rectangle is 4x6=24 square units. Now imagine a line is drawn from (0,4) to (4,0). Every ordered pair lying below this line will fulfill the inequality x+y<4. This line forms a triangle when combined with the sides of our original rectangle. The triangle has area 1/2(4)^2 = 8. If we choose any ordered pair inside the rectangle at random, there is an 8/24 = 1/3 chance that such ordered pair will lie inside the triangle we created. Hence, the probability is 1/3.
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by Max@Math Revolution » Wed Jun 15, 2016 6:45 pm
As you can see from the diagram below, the correct answer is 1/3. Hence, B is our answer choice.
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