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-   -   Geometric probability, conditional probability with area of square (https://www.askmehelpdesk.com/showthread.php?t=407720)

  • Oct 19, 2009, 03:09 PM
    SusanCher89
    geometric probability, conditional probability with area of square
    Chose a point at random in a square with sides 0<x<1 and 0<y<1. Let X be the x coordinate and Y be the y coordinate of the point chosen. Find the conditional probability P(y<1/2 / y>x).
  • Oct 20, 2009, 04:33 AM
    Unknown008

    Well, you can do it fairly easily with the help of a sketch. On graph paper, your square will have a vertice on the origin and two vertices on the x and y axes. Have dotted lines on the lines y=1 and x=1. Then, graph the line y=1/2 and y=x. I guess your '/' in your probability means 'and'. Then, find the area below the line y=1/2, which is above the line y=x.

    Below y = 1/2 because the value has to be less that 1/2.
    Above y = x because the values has to be larger than x.

    Put the resulting area over the area of the square to have your probability.
  • Mar 18, 2010, 07:25 AM
    Unknown008
    Quote:

    Originally Posted by Unknown008 View Post
    Well, you can do it fairly easily with the help of a sketch. On graph paper, your square will have a vertice on the origin and two vertices on the x and y axes. Have dotted lines on the lines y=1 and x=1. Then, graph the line y=1/2 and and y=x. I guess your '/' in your probability means 'and'. Then, find the area below the line y=1/2, which is above the line y=x.

    Below y = 1/2 because the value has to be less that 1/2.
    Above y = x because the values has to be larger than x.

    Put the resulting area over the area of the square to have your probability.

    Oops, now that I learned about conditional probabilities, I see that his is wrong.
    The way you would solve it using the method I described would give you 1/8, but the answer is actually 1/4.

    This is because:



    So, for anyone who'd like to solve it...

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