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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.

 Apr 26, 2015

Best Answer 

 #1
avatar+111438 
+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).

 

 

  

 Apr 26, 2015
 #1
avatar+111438 
+10
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

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