PROBABILITY 785 all be cut by any one straight line), the mean value of the area of the triangle XYZ is the triangle GG G", determined by the three centres of gravity of the spaces. For example Two points X, Y are taken at random within a triangle. What is the mean area M of the triangle XYC, formed by joining them with one of the angles of the triangle ? Bisect the triangle by the line CD ; let Mj be the mean value when both points fall in the triangle ACD, and M 2 the value when one falls in ACD and the other in BCD; then 2M = M 1 + M 2 . But M! = P[; and M 2 = GG C, where G, G are the centres of gravity of "ACD, BCD, this being a case of the above theorem; hence M.,= ijABC, and M-^ABC. Hence the chance that a new point Z falls on the triangle XYC is J T ; and the chance that three points X,Y,Z taken at random form, with a vertex C, a re-entrant quadrilateral, is . 67. If M be a mean value depending on the positions of n points falling on a space A ; and if this space receive a small increment a, and M be the same mean when the n points are taken on A + a, and Mj the same mean when one point falls on a and the remaining n-l on A ; then, the sum of all the cases being M (A + a) n , and this sum consisting of the cases (1) when all the points are on A, (2) when one is on o the others on A (as we may neglect all where two or more fall on a), we have .-. (M -M)A = z a (M x -M) ..... (68), as M nearly = M. As an example, suppose two points X, Y are taken in a line of length I, to find the mean value M of (XY)", as in art. 63. If I receives an increment dl, formula (68) gives Now M! here = the mean nth power of the distance of a single point taken at random in I from one extremity of I ; and this is l"(n + 1)~ l (as is shown by finding the chance of n other points falling on that distance) ; hence as in art. 63, C being evidently 0. 68. If;; is the probability of a certain condition being satisfied by the n points within A in art. 67, p the same probability when they fall on the space A + a, and p l the same when one point falls on a. and the rest on A, then, since the numbers of favourable cases are respectively _?/( A + a)", pA.", iip^A."- 1 , we find (p p}A.=na(pip) (69). Hence if ^> =_p then p=p this result is often of great value. Thus if we have to find the chance of three points within a circle forming an acute-angled triangle, by adding an infinitesimal con centric ring to the circle, we have evidently p =p ; hence the required chance is unaltered by assuming one of the three points taken on the circumference. Again, in finding the chance that four points within a triangle shall form a convex quadrilateral, adding to the triangle a small band between the base and a line parallel to it, the chance is clearly unaltered. Therefore by (69) we may take one of the points at random in the base of the triangle without altering the pro bability. 69. Historically, it would seem that the first question given on local probability, since Buffon, was the remarkable four-point problem of Prof. Sylvester. It is, in general, to find the pro bability that four points taken at random within a given boundary shall form a re-entrant quad rilateral. It is easy to see that this problem is identical with the problem of finding the mean area of the triangle formed by three points taken at random ; for, it M be this mean, and A the given area, the chance of a fourth point falling on the triangle is M/A ; and the chance of a re-entrant quadrilateral is four times this, or 4M/A. A w Let the four points be & taken within a triangle. We may take one of them W (fig. 3) at random on the base (art. 68) ; the others X, Y, Z within the triangle. Now the four lines from the vertex B to the four points are as likely to occur in any specified order as any other. Hence it is an even chance that X, Y, Z fall on one of the triangles ABW, CBW, or that two fall on one of these triangles and the remaining one on the other. Hence the probability of a re-entrant quadrilateral is ift+iPs, where p t = prob. (WXYZ re-entrant), X,Y,Z in one triangle ; p. t do., X in one triangle, Y in the other, Z in either. But PI = T> ( ap t- 66). Now to find p.,; the chance of Z falling within the triangle WXY is the mean area of WXY divided by ABC. Now by the principle in art. 66, for any particular position of W, M(WXY) = WGG , where G, G are the centres of gravity of ABW, CBW. It is easy to see that W T GG = 1 & ABC = ? putting ABC = 1. Now, if Z falls in CBW, the chance of WXYZ re- entrant is 2M(IYW), for Y is as likely to fall in WXZ as Z to fall in WXY ; also if Z falls in ABW the chance of WXYZ re-entrant is 2M(IXW). Thus the whole chance is j 2 = 2M(IYW + IXW) = . Hence the probability of a re-entrant quadrilateral is 1 . 4 4-1 . ? 1 2 IfT 5 u 3 That of its being convex is f. 70. If three points X, Y, Z are taken at random in a triangle, the mean value of the triangle XYZ = -jVf the given triangle. For we have seen that the chance of four points forming a re-entrant figure is 4M/A, where M is the required mean and A the given triangle ; as this has been shown to be ^, M = f a A. 71. Let the three points be taken within a circle; and let M be the mean value of the triangle formed. Adding a concentric ring o, we have (68) since M :M as the areas of the circles, M = r M. where Mj is the value of M when one of the points is on the circumference. Take fixed ; we have to find the mean value of OXY (fig. 4). Taking (p, 6} (p, 6 ) as coordinates of X, Y, . (OXY) . o sin(0 - e )pp dpdp d6dO = (7r 2 a 4 )- 1 . ffrW*. sin(0- 6 )dedO , putting r = OH, ?- = OK ; asr = 2asin0, / = 2f sin , M, = 4-.. ( ^ f*f 9 A-D*9wMKD.(9-tf)Mdff. 7T-0. 4 9 JQ JQ P-ofessor Sylvester has remarked that this double integral, by means of the theorem " rx f( x , y}dxdy=f a f X f(a-y,a-xy. X dy , is easily shown to be identical with TT /~0 sin 4 Osin s Q cosddddO =J 35a 2 . . M 35 Hence the probability that four points within a circle shall form a re-entrant figure is 35 P 127T 2 72. Professor Sylvester has remarked that it would be a novel question in the calculus of variations to determine the form of the convex contour which renders the probability a maxi mum or minimum that four points taken within it shall give a re-entrant quadri lateral. It will not be difficult to show, by means of the principles we have been examining, that the circle is the contour which gives the minimum. For, if p be the probability of a re entrant figure for four points within a circle of area A, p the same probability when a small addition a, of any kind p. , which still leaves the whole contour convex, is made to the circle, we have by (69) where p l -=te probability when one point is taken in a that is, in the limit, when one point is taken on the circumference of the circle. But Pi=p, as is shown in art. 68 ; hence p - p = . Hence any infinitesimal variation of the contour from the circum ference of the circle gives Sp, the variation of the probability, zero, the same method being applicable when portions are taken away, instead of being added, provided the contour is left convex.
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