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 $A$?
+14903 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
\(probability=\frac{\pi r^2}{6}:\\(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi r^2}{6})\)
\(probability=\frac{\pi }{6}:(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi}{6})\) [ \(\times\frac{6}{\pi}\)
\(probability=1:(\frac{a^2\cdot 6 \cdot\sqrt{3}}{4\pi}-1)\)
\(prohability=1:(\frac{a^2\cdot 3 \cdot\sqrt{3}}{2\pi}-1)\)
\(prohablity=1:(0.827a^2-1)\)
!
+9465 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.
+128406 Yours is correct, hectictar.........( target area of triamgle) / (whole area of triangle) = desired probability
+14903 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.
+9465 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 \(\frac38\))
+128406 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
