+0

# 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 exce

0
313
1
+1773

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
+83935
+10

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
Sort:

#1
+83935
+10

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

### 9 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details