Question

Three points are taken at random on the circumference of a circle. What is the probability that they will all lie in a semicircle?

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{3}{4}\)
• D. \(\frac{2}{3}\)

Hint:

For \(N\) random points on a circle, the general formula for the probability that they all lie within some semicircle is:
\[ P = \frac{N}{2^{N-1}} \]
For \(N = 3\), we get:
\[ P = \frac{3}{2^{3-1}} = \frac{3}{4} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a classic problem in geometric probability.
We want to find the probability that three randomly chosen points on the circumference of a circle can be enclosed within some semicircle (an arc of length \(\le \pi\) radians or \(180^\circ\)).

Step 2: Detailed Explanation:
Let the three points chosen at random on the circle be \(A\), \(B\), and \(C\).
Because of the circular symmetry, we can fix the position of the first point \(A\) without loss of generality.
The positions of the other two points, \(B\) and \(C\), can be defined by their clockwise angular distances from \(A\), denoted as \(\theta_1\) and \(\theta_2\) respectively, where \(0 \le \theta_1, \theta_2 \le 2\pi\).
The sample space of possible angles is a square of side \(2\pi\) in the \(\theta_1\)-\(\theta_2\) plane, with an area of:
\[ \text{Area}(S) = (2\pi) \cdot (2\pi) = 4\pi^2 \]
For all three points to lie in some semicircle, the maximum angular distance between any two points must be less than or equal to \(\pi\).
This condition is violated only if the points cannot be contained in any semicircle, which happens if the angular distances partition the circle such that all three intervals between the points are less than \(\pi\) (creating a triangle that contains the center of the circle).
Let us find the probability of this opposing event (that they do not lie in any semicircle):
This occurs if:
1. \(\theta_1 < \pi\) and \(\theta_2 > \theta_1 + \pi\)
2. \(\theta_2 < \pi\) and \(\theta_1 > \theta_2 + \pi\)
The combined area of these unfavorable regions in the square is \(\frac{1}{4}\) of the total area of the square.
Therefore, the probability that the center of the circle is inside the triangle formed by the three points is:
\[ P(\text{not in a semicircle}) = \frac{1}{4} \]
Using the complement rule, the probability that the three points all lie in some semicircle is:
\[ P(\text{all in a semicircle}) = 1 - P(\text{not in a semicircle}) = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 3: Final Answer:
The correct option is (C).