![]() ![]() Then we select a point at random on it, and draw the chord through the point and normal to the diameter. (2) Choose the chord centre-point on a fixed diameter. Thus, the probability of this event is ⅓. point A ) and exceeds it for points on the centre arc (e.g., point B ). The chord length is less than the side of the triangle for the arcs adjacent to the first endpoint (e.g. Since the other endpoint (e.g., A or B ) must be on one of three arcs, all of equal length, we may argue that each arc is equally probable. On the basis of symmetry, we may choose one endpoint of the chord to coincide with a vertex of the equilateral triangle. We will find that the probability differs for the three methods of choice. We will consider three ways of drawing a chord in the outer circle:Ĭhoose the chord centre on a fixed diameter. What is the probability that the chord length is greater than the side of an equilateral triangle inscribed in the circle? Problem: Given a circle, a chord is drawn at random. It is a simple geometric result that the radius of the outer circle is twice that of the inner one. Let’s start with an equilateral triangle and add an inscribed circle and a circumscribed circle. We will have a look at Bertrand’s Paradox (1889), a simple result in geometric probability. ![]() Many of these have been reviewed in a recent book by Prakash Gorroochurn. ![]() The history of probability theory has been influenced strongly by paradoxes, results that seem to defy intuition. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |