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Finding probability of two numbers which satisfies an inequality

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Two numbers x and y are selected from a closed interval [0,4]. To find the probability that the two numbers satisfies the condition that y[itex]^{2}[/itex][itex]\leq[/itex] x.


    2. The attempt at a solution

    Don't have any idea
     
  2. jcsd
  3. Nov 10, 2012 #2

    HallsofIvy

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    Draw a graph. First draw the parabola representing the function [itex]x= y^2[/itex], then draw the four line segments x= 0, y= 0, x= 4, y= 4 making a square, with vertices (0, 0), (16, 0), (16, 16), and (0, 16), and having the graph [itex]x= y^2[/itex], which is the same as [itex]y= x^{1/2}[/itex], crossing the square from (0, 0) to (4, 2). The set of points such that [itex]y^2\le x[/itex] with x and y from [0, 4] is the set of point below that graph. Assuming all values of x and y between 0 and 4 are "equally likely, then all points in the square are "equally likely" and the probability a point is below the parabola is the ratio of the area under the parabola to the area of the square. Find that area by integrating [itex]x^{1/2}[/itex] from x= 0 to x= 4 and then divide by the area of the square, 16.
     
  4. Nov 10, 2012 #3
    Gotcha..thanks for the help
     
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