Two numbers between $0$ and $1$ on a number line are to be chosen at random. What is the probability that the second number chosen will exceed the first number chosen by a distance greater than $\frac 14$ unit on the number line? Express your answer as a common fraction.

Mellie Apr 26, 2015

#1**+13 **

Here's the way I approached this one.......

In the square BEDA, the diagonal AE represents all the x = y values from 0 to 1. The segment FG will hold all the y values that are .25 greater than their associated x values at any point on this segment. So, since the square holds all possible (x,y) values from 0 to 1 where x is the first number chosen and y is the second one chosen, the only y values of interest lie in the triangle GBF. All the y values in here will exceed all their associated x values by more than .25. Note that the x value at H = 7.5. And, at point G, the associated y value will be the largest possible, i.e., 1.

And the area of this truangle = (1/2)(.75)(.75)sin90 = .28125 = 9/32

So, the total area of BEDA encompasses all possible (x,y) values from 0 to 1. And the area of this square is just 1.

And triangle GBF is 9/32 of this area = probability that the second number chosen will exceed the first by more than .25 ..... (1/4).

CPhill Apr 26, 2015

#1**+13 **

Best Answer

Here's the way I approached this one.......

In the square BEDA, the diagonal AE represents all the x = y values from 0 to 1. The segment FG will hold all the y values that are .25 greater than their associated x values at any point on this segment. So, since the square holds all possible (x,y) values from 0 to 1 where x is the first number chosen and y is the second one chosen, the only y values of interest lie in the triangle GBF. All the y values in here will exceed all their associated x values by more than .25. Note that the x value at H = 7.5. And, at point G, the associated y value will be the largest possible, i.e., 1.

And the area of this truangle = (1/2)(.75)(.75)sin90 = .28125 = 9/32

So, the total area of BEDA encompasses all possible (x,y) values from 0 to 1. And the area of this square is just 1.

And triangle GBF is 9/32 of this area = probability that the second number chosen will exceed the first by more than .25 ..... (1/4).

CPhill Apr 26, 2015