**Bertrand Paradox. **

We are given a circle C of radius r and we wish to determine the probability p that the length l of a “randomly selected” chord AB is greater than the length of the inscribed equilateral triangle.

We shall show that this problem can be given at least three reasonable solutions

**I. **If the center M of the chord AB lies inside the circle of radius shown in figure a (attached), then . It is reasonable, therefore, to consider as favourable outcomes all points inside the circle and as possible outcomes all points inside the circle C. Using as measure of their numbers the corresponding areas and , we conclude that

**II. **We now assume that the end A of the chord AB is fixed. This reduces the number of possibilities, but it has no effect on the value of p because the number of favourable locations of B is reduced proportionately. If B is on the 120 degree arc of DBE of figure b (attached), then . The favourable outcomes are now the points on this arc and the total outcomes all points on the circumference of the circle C. Using as their measurements, the corresponding lengths and , we obtain

.

**III. **We assume finally that the direction of AB is perpendicular to the line FK of figure c (attached). As in II, the restriction has no effect on the value of p. If the center M of AB is between G and H, then . Favourable outcomes are now the points on GH and possible outcomes all points on FK. Using as their measures the respective lengths r and , we obtain

.

We have thus found not one but three different solutions for the same problem! One might remark that these solutions correspond to three different experiments. This is true but not obvious and, in any case, it demonstrates the ambiguities associated with the classical definition, and the need for a clear specification of the outcomes of an experiment and the meaning of the terms “possible” and “favourable.”

More later,

Nalin Pithwa

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