ON INVERSE PROBABILITIES. O'l 



to find the probability which it affords to the suppo- 

 sition that a coming event shall be B, find the proba- 

 bility which A gives to every possible preceding state ; 

 multiply each probability thus obtained by the chance 

 which B would have from that state, and add the 

 results together. 



PROBLEM. There is a, lottery of 10 balls, each one 

 white or black, but which is not known : drawings are 

 made, after each of which the hall is replaced. The 

 first five drawings are white ; what chance is there 

 that the next two drawings shall be white ? 



Let S (20) denote the sum of all numbers up to 20 ; 

 S (20 2 ) the sum of the squares of all numbers up to 

 20 2 or 400 ; and so on. The possible preceding states 

 are 



(1 W, <>B) (2 W,8 B.) ---- (10 W,OB); 

 and the probabilities of W five times running from 

 each, are 



TO iV ' TV * To * To To" ' To ' T o" * Fo To> & C> 



iS iS> or 1* the 



. . -- 



certain, if the last state existed. The numerators of 

 these products (the common denominator being 10 5 ,) 

 are I 5 , 2 5 , - 10 5 : whence, page 55., the proba- 

 bilities of the several states are 



is & ......... 1Q5 



S 105' S 105* S 105 



By the same reasoning, the probabilities of the proposed 

 events (two more white balls,) are 



12 22 ^ ^ 1Q2 



10 2 10 2 102 ; 



the different preceding states being successively sup- 

 posed to exist; whence the actual chance which the 

 observed event gives to the proposed is 



_I!_x.il+_^Lx^l + ...... * 105 .' 2 



S 105 102 S 105 102 S 105 102 ; 

 S 107 



which is 



