# Geometric probability.

• December 19th 2010, 03:18 PM
nikki712
Geometric probability.
Please see attachment. He's stuck on how to figure out the last problem (the one on the right side of the attachment). I'm not sure how to help him. Can someone please help me?

Attachment 20159
• December 19th 2010, 03:23 PM
dwsmith
The square is square so all sides are equal.

If we divide it horizontally, we have two smaller squares (left side) which are half shaded.

So we 50% shaded right and 1/2 of 50% shaded = 75%

Area of a circle = $\pi r^2$ and r = 7 so the area of our circle is $\pi 7^2=49\pi$

The diagonal of the square is 2*7=14 and the sides are equal. $c^2=a^2+b^2\Rightarrow c^2=a^2+a^2\Rightarrow c^2=2a^2$

c is the diagonal so c =14.

$14^2=2a^2\Rightarrow 196=2a^2\Rightarrow \mbox{divide by 2} \ \ 98=a^2$

Area of a square is a*a so we have 98 is the area.

Area of shaded region is $49\pi-98=49(\pi-2)$

Now divide by total area. $\displaystyle\frac{49(\pi-2)}{49\pi}=\frac{\pi-2}{\pi}$
• December 19th 2010, 03:38 PM
Quote:

Originally Posted by nikki712
Please see attachment. He's stuck on how to figure out the last problem (the one on the right side of the attachment). I'm not sure how to help him. Can someone please help me?

Attachment 20159

The probability of a point being in the shaded area is

$\displaystyle\frac{shaded\;region\;area}{circle\;a rea}$

Use Pythagoras' theorem to calculate half the length of the side of the square.

$\displaystyle\ x^2+x^2=7^2\Rightarrow\ 2x^2=7^2\Rightarrow\ x=\frac{7}{\sqrt{2}}$

$\displaystyle\ 2x=square\;side\;length=\frac{(2)7}{\sqrt{2}}=\fra c{7\sqrt{2}\sqrt{2}}{\sqrt{2}}=7\sqrt{2}$

Square area is $(2x)^2=(2)7^2$

The circle area is ${\pi}7^2$

Finally calculate the probability of a random point being in the shaded region.

$\displaystyle\ P=\frac{{\pi}7^2-(2)7^2}{{\pi}7^2}=\frac{{\pi}-2}{{\pi}}$
• December 19th 2010, 03:44 PM
nikki712
Thanks!!!

Can anyone tell me if the problem on the left side of the picture is right? Just wanna make sure. :)
• December 19th 2010, 03:45 PM
dwsmith
Probability: is 75% for the first one and $\displaystyle \frac{\pi-2}{\pi}$ for the second one.
• December 19th 2010, 03:47 PM
Quote:

Originally Posted by nikki712
Thanks. Now is this answer the area, or probability? He needs to find the probability.

Yes, the probability is the shaded region divided by the circle area.

The circle contains the total area.
The shaded region is the "difference" between the circle area and the square area.
Hence the square area is subtracted from the circle area to obtain the numerator of the probability fraction.
• December 19th 2010, 03:49 PM
dwsmith
Quote:

Originally Posted by nikki712
Thanks!!!

Can anyone tell me if the problem on the left side of the picture is right? Just wanna make sure. :)

5/6 isn't correct for either question.
• December 19th 2010, 04:40 PM
nikki712
Well that sucks. I will tell him to solve the problem the same way you guys solved the other problem. Thanks for your help!!!
• December 20th 2010, 12:18 AM
Opalg
Quote:

Originally Posted by nikki712
Well that sucks. I will tell him to solve the problem the same way you guys solved the other problem. Thanks for your help!!!

The reason that 5/6 is wrong is that the six triangles are not all the same size. The two triangles in the central diamond are larger than the four triangles around the outside, in fact their area is twice as large.
• December 20th 2010, 06:36 AM
Soroban
Hello, nikki712!

Another approach . . .

Quote:

A point is randomly chosen in the circle.
Find the probability that the point is in the shaded region.

Code:

* * *
*:::::::::::*
o - - - - - - - o
*|              |*
:|              |:
*:|              |:*
*:|      *      |:*
*:|        * 7  |:*
:|          *  |:
*|            * |*
o - - - - - - - o
*:::::::::::*
* * *

The area of the circle is: . $A_c \:=\:\pi r^2 \:=\:\pi(7^2) \:=\:49\pi$

The square is a rhombus.
The area of a rhombus is one-half the product of its diagonals.
. . $A_s \:=\:\frac{1}{2}(14)(14) \:=\:98$

Hence: . $P(\text{square}) \:=\:\dfrac{98}{49\pi} \:=\:\dfrac{2}{\pi}$

Therefore: . $P(\text{not-square}) \:=\:1 - \dfrac{2}{\pi} \:=\:\dfrac{\pi-2}{\pi}$

• December 20th 2010, 09:16 AM
Soroban
Hello, nikki712!

For the first problem, draw a horizontal line through the center of the square.

Code:

*-------*-------*
|:::::* |:*:::::|
|:::*  |:::*:::|
|:*    |:::::*:|
* - - - +:-:-:-:*
|:*    |:::::*:|
|:::*  |:::*:::|
|:::::* |:*:::::|
*-------*-------*

The square is divided into eight congruent right triangles.