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Math Help - Geometric probability.

  1. #1
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    Geometric probability.

    Half of a circle is inside a square and half is outside, as shown. If a point is selected at random inside the square, find the probability that the point is not inside the circle. (r is the radius)



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    ~Vince
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  2. #2
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    Quote Originally Posted by vince5002001 View Post
    Half of a circle is inside a square and half is outside, as shown. If a point is selected at random inside the square, find the probability that the point is not inside the circle. (r is the radius)



    TYA

    ~Vince
    You know that:

    The area of the square = (2r)^2 \longleftrightarrow 4r^2
    You know that the area of the circle = \pi r^2
    So the part of the circle that is IN the square = \frac {\pi r^2}{2}
    So (4r^2 - \frac {\pi r^2}{2}) gives you the area of the part of the square that isn't part of the half-circle.

    Can you do the rest?
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  3. #3
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    - = (4r^2)2 = 8r^2 -\pi r^2

    Then would I divide by r^2? and get 8-\pi then substitute 8 for the area of the square?

    Giving you  {8-\pi}/{8} for the probability that the point is not inside the circle but inside the square.
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  4. #4
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    Quote Originally Posted by vince5002001 View Post
    - = (4r^2)2 = 8r^2 -\pi r^2

    Then would I divide by r^2? and get 8-\pi then substitute 8 for the area of the square?

    Giving you  {8-\pi}/{8} for the probability that the point is not inside the circle but inside the square.
    Factorizing (4r^2) - \frac{\pi r^2}{2} gives me the following:

    \frac {2(4r^2)}{2} - \frac {\pi r^2}{2} \longleftrightarrow \frac {2(4r^2)-\pi r^2}{2} \longleftrightarrow \frac {8r^2-\pi r^2}{2}.

    We know that the area of the half-circle divided by the area of the whole square multiplied by 100 gives us the percentage that the half-circles occupies in the square. Then, 100% minus the percentage found gives us the percentage of the rest of the square, which is the probability we're looking for. But I've tried to find a concrete percentage but I couldn't, I'm a bit rusty tonight and it's getting late Tell me if you get anything new...

    Those are the calculations for what I've stated above:

    \frac {2\pi r^2}{8r^2 - \pi r^2} x 100

    \frac {2\pi r^2}{r^2(8-\pi)} x 100

    \frac {2\pi}{(8-\pi)} x 100

    \frac {200\pi}{(8-\pi)}% = The percentage that the half-square occupies

    100% - \frac {200\pi}{(8-\pi)}%= The probability we're looking for...

    \frac {200\pi}{(8-\pi)} \approx 129%

    100% - 129% = -29%
    Last edited by Jameson; May 7th 2009 at 07:36 PM.
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  5. #5
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    Quote Originally Posted by Sodapop View Post
    Factorizing (4r^2) - \frac{\pi r^2}{2} gives me the following:

    \frac {2(4r^2)}{2} - \frac {\pi r^2}{2} \longleftrightarrow \frac {2(4r^2)-\pi r^2}{2} \longleftrightarrow \frac {8r^2-\pi r^2}{2}.

    We know that the area of the half-circle divided by the area of the whole square multiplied by 100 gives us the percentage that the half-circles occupies in the square. Then, 100% minus the percentage found gives us the percentage of the rest of the square, which is the probability we're looking for. But I've tried to find a concrete percentage but I couldn't, I'm a bit rusty tonight and it's getting late Tell me if you get anything new...

    Those are the calculations for what I've stated above:

    \frac {2\pi r^2}{8r^2 - \pi r^2} x 100

    \frac {2\pi r^2}{r^2(8-\pi)} x 100

    \frac {2\pi}{(8-\pi)} x 100

    \frac {200\pi}{(8-\pi)}% = The percentage that the half-square occupies

    100% - \frac {200\pi}{(8-\pi)}%= The probability we're looking for...

    \frac {200\pi}{(8-\pi)} \approx 129%

    100% - 129% = -29%
    This calculation is wrong. A probability cannot be negative.

    The correct answer is \frac{4r^2 - \frac{1}{2} \pi r^2}{4r^2} which simplifies to 1 - \frac{\pi}{8}.
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