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.




 Jan 5, 2021

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).

 Jan 5, 2021

It was wrong but I appreciate the effort to answer it smiley

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

HelpPls123abc  Jan 5, 2021
edited by HelpPls123abc  Jan 5, 2021

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


cool cool cool

 Jan 5, 2021
edited by CPhill  Jan 5, 2021

TYSM CPHILL that was helpful and correct :D

HelpPls123abc  Jan 6, 2021

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