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 in some semicircle is:
\[ P = \frac{N}{2^{N-1}} \]
For \(N = 3\) points:
\[ P = \frac{3}{2^2} = \frac{3}{4} \]

Answer 1

The Correct Option is C

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

Step 2: Detailed Explanation:
Let the three points be \(A\), \(B\), and \(C\).
Because of circular symmetry, we can fix the position of the first point \(A\) on the circumference of the circle.
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 in the \(\theta_1\)-\(\theta_2\) plane with side length \(2\pi\), meaning the total area of the sample space is:
\[ \text{Area}(S) = (2\pi) \times (2\pi) = 4\pi^2 \]
For the three points to lie in a semicircle, the maximum angular distance between any two of them must be less than \(\pi\) (or \(180^\circ\)).
The only way they *do not* lie in a semicircle is if they partition the circle such that all three intervals between the points are less than \(\pi\) (which means the triangle formed by \(A, B, C\) contains the center of the circle).
This condition of NOT lying in a semicircle is met 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 two unfavorable regions in the square is \(\frac{1}{4}\) of the total area of the square.
Therefore, the probability that the three points do not lie in a semicircle is:
\[ P(\text{unfavorable}) = \frac{1}{4} \]
The probability that they all lie in a semicircle is the complement:
\[ P = 1 - P(\text{unfavorable}) = 1 - \frac{1}{4} = \frac{3}{4} \]

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