# Geometric probability.

• May 7th 2009, 02:58 PM
vince5002001
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)

http://i334.photobucket.com/albums/m...alfacricle.jpg

TYA

~Vince
• May 7th 2009, 03:43 PM
Sodapop
Quote:

Originally Posted by vince5002001
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)

http://i334.photobucket.com/albums/m...alfacricle.jpg

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?
• May 7th 2009, 04:50 PM
vince5002001
http://www.mathhelpforum.com/math-he...8bcce9ed-1.gif - http://www.mathhelpforum.com/math-he...aed795f9-1.gif = $(4r^2)2 = 8r^2 -\pi r^2$

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.
• May 7th 2009, 05:09 PM
Sodapop
Quote:

Originally Posted by vince5002001
http://www.mathhelpforum.com/math-he...8bcce9ed-1.gif - http://www.mathhelpforum.com/math-he...aed795f9-1.gif = $(4r^2)2 = 8r^2 -\pi r^2$

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 :p 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%
• May 8th 2009, 03:10 AM
mr fantastic
Quote:

Originally Posted by Sodapop
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 :p 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 $1 - \frac{\pi}{8}$.