Two numbers between 0 and 1 on a number line are chosen at random. What is the probability that the second number chosen will: exceed the first number chosen? Express your answer as a common fraction.

TY

HelpPls123abc Jan 5, 2021

#1**-1 **

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).

hihihi Jan 5, 2021

#2**0 **

It was wrong but I appreciate the effort to answer it

I think that was the answer for a different version of the problem.

HelpPls123abc
Jan 5, 2021

#3**+3 **

Here's my attempt

See the following image :

There are two regions....A and B

The diagonal line y = x comtains all the values where the two values chosen are equal

Region B contains all the points where the first coordinate is greater than the second

In region A, the second coordinate of every point is greater than the first coordinate.....so the probability of landing in this area = 1/2

CPhill Jan 5, 2021