Hi guys,
Long time no see. I saw a question looks something like this:
Get three points randomly on a circle and connect them to draw a triangle. What's the probability for this triangle to include the center of the circle?
The answer was said to be 0.25 but my calculation shows the answer is $$\pi/6$$.
Here is how I calculated. Please let me know if you observe an error. Thanks
Step1: Take a random point(A) on the circle. The probability P1=1.
Step2: Take the second point B on the circle and say the center is point O. The degree of AOB is X and 0<X<= pi. The probability of getting this point is P2= 2 * X/ (2pi) = X/pi
Step3: To get the point C making triangle ABC to include the center, we need to extend the line AO and BO respectively and intersect on the circle at A' & B'. Point C must fall on the curve A'B' and the probability of this is P3=X/(2pi).
Step4: P = P1*P2*P3 = X^2 / (2 pi^2). Do an integral $$\int_{0}^{\pi } x^2 /2 \pi^2 = \pi / 6$$
One probability question
Re: One probability question
Oh. Something was wrong. P2 will be 0.Hom wrote:Hi guys,
Long time no see. I saw a question looks something like this:
Get three points randomly on a circle and connect them to draw a triangle. What's the probability for this triangle to include the center of the circle?
The answer was said to be 0.25 but my calculation shows the answer is $$\pi/6$$.
Here is how I calculated. Please let me know if you observe an error. Thanks
Step1: Take a random point(A) on the circle. The probability P1=1.
Step2: Take the second point B on the circle and say the center is point O. The degree of AOB is X and 0<X<= pi. The probability of getting this point is P2= 2 * X/ (2pi) = X/pi
Step3: To get the point C making triangle ABC to include the center, we need to extend the line AO and BO respectively and intersect on the circle at A' & B'. Point C must fall on the curve A'B' and the probability of this is P3=X/(2pi).
Step4: P = P1*P2*P3 = X^2 / (2 pi^2). Do an integral $$\int_{0}^{\pi } x^2 /2 \pi^2 = \pi / 6$$

 Posts: 5
 Joined: Fri Dec 30, 2011 2:50 am
Re: One probability question
You are on the right track! With the first point fixed, the second point will be at an angle uniformly distributed in [0, pi] with respect to the center. The probability of a third point landing such that the triangle contains the center for a given angle $$X$$ is just $$X / 2\pi$$, as you surmised by extending the chords.
So integrating the distribution and normalizing it (since it is 0 to pi instead of 0 to 1) gives
$$\frac{1}{\pi}\int_0^\pi{\frac{X}{2\pi} dX} = \frac{1}{4}$$
So integrating the distribution and normalizing it (since it is 0 to pi instead of 0 to 1) gives
$$\frac{1}{\pi}\int_0^\pi{\frac{X}{2\pi} dX} = \frac{1}{4}$$