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 $x < y$?

Mellie
Apr 27, 2015

#3**+10 **

You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... )

If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values.

Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2.

geno3141
Apr 27, 2015

#1**+5 **

Of the four points: (0,0), (4,0), (4,1), and (0,1) only the point (0,1) has an x-value smaller than the y-value.

The probability that the x-value is less than the y-value is 1/4.

geno3141
Apr 27, 2015

#2**+5 **

Thank you geno, but this was incorrect. Sorry!!!!!!

$$The point $(x,y)$ satisfies $x < y$ if and only if it belongs to the shaded triangle bounded by the lines $x=y$, $y=1$, and $x=0$, the area of which is 1/2. The ratio of the area of the triangle to the area of the rectangle is $\frac{1/2}{4} = \boxed{\frac{1}{8}}$.

[asy]

draw((-1,0)--(5,0),Arrow);

draw((0,-1)--(0,2),Arrow);

for (int i=1; i<5; ++i) {

draw((i,-0.3)--(i,0.3));

}

fill((0,0)--(0,1)--(1,1)--cycle,gray(0.7));

draw((-0.3,1)--(0.3,1));

draw((4,0)--(4,1)--(0,1),linewidth(0.7));

draw((-0.5,-0.5)--(1.8,1.8),dashed);

[/asy]$$

Mellie
Apr 27, 2015

#3**+10 **

Best Answer

You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... )

If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values.

Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2.

geno3141
Apr 27, 2015

#4**+5 **

I get something a little different here than geno.....see the following pic........

All points inside triangle ADF will have x values less than their associated y values. And the area of this triangle = 1/2 sq units

And the area of the whole rectangle = (4)(1) = 4 sq units

So....the probability that a random point has an x value less than its y value = (1/2) / 4 = 1/8

CPhill
Apr 28, 2015