The random endpoints method entails choosing two random points on the circumference and joining them to form a chord of the circle. Now, the triangle can be rotated so that one of these points coincides with one of its vertices. As visible in figure 1, a chord that goes through the angle at that vertex is longer than a side length whilst a chord that doesn’t is shorter. Since said angle is necessarily 60 ° , the probability of the chord exceeding the sides in length is 60 180 - or more simply, 1 3 . Notice that this is equivalent to picking an angle in the range 0 – 180 that the chord will form with a tangent.
The random radius method requires for a radius of the circle to be drawn and the triangle rotated such that the radius is perpendicular to one of the sides of the triangle. Next, a random chord parallel to the side is to be drawn through the radius. From figure 2 we can see that such a chord is longer than the side if the point where it meets the radius is closer to the centre than the point where the radius intersects the side. The side of the triangle bisects the radius exactly so this corresponds to 1 2 of such chords.
Finally, the random midpoint method involves the following procedure: a
randomly selected point within the circle is selected and the chord that has said point as its midpoint is drawn through the circle. A concentric circle with 1 2 the radius should be drawn as shown in figure 3. If the chosen point lies within the smaller circle, then the chord drawn through it is longer than a side of the triangle. The area of the smaller circle is 1 4 that of the larger circle, which means that the probability of the chord drawn being longer than a side length of the triangle is 1 4 .
Furthermore, a similar problem can be found in ‘Fifty Challenging Problems In Probability’ (Frederick Mosteller, 1965), in which the question asks ‘if a chord is randomly drawn in a circle, what is the probability that its length will exceed the radius?’. This particular problem also has three solutions.
The first method: ‘Assume that the distance of the chord from the centre of the circle is evenly (uniformly) distributed from 0 to r. Since a regular hexagon of side r can be inscribed in a circle, to get the probability, merely find the distance d from the center and divide by radius. Note that this is the altitude of an equilateral triangle of side r. Therefore from plane geometry we get d = √ 2 − 2 4 = √3 2 and consequently, the desired probability is approx. = 0.866’ The second method: ‘Assume that the midpoint of the chord is evenly distributed over the interior of the circle. Consulting the figure again, we see that the chord is longer than the radius when the midpoint of
the chord is within d of the centre. Thus all points in the circle of radius d, concentric with the original circle, can serve as midpoints of the chord. Their fraction, relative to the area of the original circle, is 𝜋 2 𝜋 2 =
2 2
= 3 4 = 0.75. This probability is the square of the result we got in method 1.
The third method: ‘Assume that the chord is determined by two points chosen so that their positions are independently evenly distributed over the circumference of the original circle. Suppose that the first point falls at A in this figure. Then, for the chord to be shorter than the radius, the second point must fall on the arc ABC, whose length is 1 3 the circumference. Consequently, the probability that the chord is longer than the radius is 1- 1 3 = 2 3 .’
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