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Triangle $ABC$ is equilateral with side length of 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?

Guest Apr 14, 2017
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#1
+6722
+3

Triangle $ABC$ is equilateral with side length of 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?

Within the circle section around A, X is not more than 1 unit from A.

The probability sought is equal to the ratio of the area of the circular segment to the rest of the area of the triangle.

The side of the triangle is a. r = 1

$$probability=\frac{\pi r^2}{6}:\\(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi r^2}{6})$$

$$probability=\frac{\pi }{6}:(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi}{6})$$              [   $$\times\frac{6}{\pi}$$

$$probability=1:(\frac{a^2\cdot 6 \cdot\sqrt{3}}{4\pi}-1)$$

$$prohability=1:(\frac{a^2\cdot 3 \cdot\sqrt{3}}{2\pi}-1)$$

$$prohablity=1:(0.827a^2-1)$$

!

asinus  Apr 14, 2017
edited by asinus  Apr 14, 2017
edited by asinus  Apr 14, 2017
#2
+4174
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I didn't have any idea how to do this until I read asinus's answer!

But after I saw his I wanted to try it.

Here, A is the center of the circle with a radius of 1.

probability that X ≤ 1 unit away from A = $$\frac{\text{area of blue sector}}{\text{area of triangle}}$$

$$\frac{\text{area of blue sector}}{\pi*1^2}=\frac{60^{\circ}}{360^{\circ}} \rightarrow \text{area of blue sector}=\frac{\pi}{6}$$

$$\text{area of triangle}=(2)*(\frac12)*(\frac32)*(\frac{3\sqrt3}{2}) =\frac{9\sqrt3}{4}$$

$$\text{probability}=\frac{\frac{\pi}{6}}{\frac{9\sqrt3}{4}}=\frac{\pi}6\cdot\frac4{9\sqrt3}=\frac{4\pi}{54\sqrt3}=\frac{2\sqrt3\pi}{81}$$

My answer is a little different because I put the area of the whole triangle in the denominator, but asinus put the area of the rest of the triangle, not including the sector, in the denominator. I don't really know for sure which is correct.

hectictar  Apr 14, 2017
#3
+75376
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Yours is correct, hectictar.........( target area of triamgle) / (whole area of triangle) = desired probability

CPhill  Apr 14, 2017
#4
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The point X meets randomly from above on two surfaces of equal importance:

the circular section$$(\leq 1 \ unit\ from\ vertex\ A)$$ and the rest of the triangle$$(>1 \ unit\ from\ vertex\ A)$$.
The probability can only be the quotient of these two surfaces.

asinus  Apr 14, 2017
edited by asinus  Apr 14, 2017
edited by asinus  Apr 14, 2017
edited by asinus  Apr 14, 2017
#5
+4174
+2

Hmm.. this is my (very limited!) understanding of probability...

$$\text{probability}=\frac{\text{number of ways something can happen}}{\text{total number of possible outcomes}}$$

So... if you have a bag of 3 blue marbles and 8 red marbles,

the probability of drawing a blue marble = $$\frac{3}{11}$$ (and not $$\frac38$$)

hectictar  Apr 14, 2017
edited by hectictar  Apr 14, 2017
#6
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0

We compare two surfaces.
The sum of the two surfaces is not interesting (given the given question).

asinus  Apr 15, 2017
#7
+75376
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Asinus....you have given the odds that   "x"  is no  more than 1 unit away from "A,"  not the probability

To see the difference .....look at the following image :

The probability that we choose the green area in the above circle  =

[ area that is green] / [ area that is green + area that is not green ]  =

[1/4 of circle] / [ 1/4 of circle + 3/4 of circle] =

[1/4 of circle] / [ whole circle]   =    1/4

But the odds that we choose the green area  =

[area that is green] / [area that is not green]  =

[1/4 of circle] / [ 3/4 of the circle] =    1/3   =  1 : 3

CPhill  Apr 15, 2017
#8
+6722
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Thank you Chris, you have explained to me well.
I have looked at chance and probability as the same.
Sometimes it is not easy for me to understand English properly.
You surely look at this.

Greeting asinus: -)

asinus  Apr 15, 2017

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