ON INVERSE PROBABILITIES. 65 



The preceding affords some view of the way in 

 which chances are obtained, in cases where the ante- 

 cedent probability of the events stated may be any 

 whatever. The following are conclusions upon the same 

 subject, obtained by a more complicated reasoning of 

 the same kind. 



If an event, each repetition of which may be either 

 A or B, have happened m-\-n times, and if A have 

 occurred m times, and B n times ; then it is m -\- 1 

 to n + 1 that the next event shall produce A, and 

 not B. And in the same case the chance that out of 

 p -f- q events to come, p shall produce A, and q shall pro- 

 duce B, is (see pages 15 and 16, for explanation of [ ]). 

 \p+q\ 



[f] x M x [m+n+2> m+n+p+q+ 1]' 

 and 



. 

 [m + n+2, m + n+p + 1] [m + n+2, m+n+q + l] 



are the chances that in p new events, all shall give A 

 and that in q new events all shall give B. 



EXAMPLE. In a lottery containing an unlimited 

 number of balls, in which the proportion of black and 

 white is absolutely unknown, six drawings give four 

 white and two black ; what are the chances that four 

 drawings more shall give all white, or one only black, 

 or two only black, &c. 



Let us first take the case of three white and one black : 

 here m = 4, n = 2, p = 3, q =. 1. 



[p + q] = 1.2.3.4, [m + l,m+p] = 5.6.7, 

 [n + l,n + q] =3[p] = 1.2.3. [q] = 1, 

 [m + w + 2, m + n+p + q + l~\ = 8. 9. 10.11. 



Chance required is 1-2.3.4. x 5.6.7. * 3. _ 7 

 1.2.3, x 1. x 8.9.10.11. 22 



That two shall be white and two black (m = 4, n = % 3 

 p = 2, q = 2), the chance is 



1 2.3.4. x5.6. x3.4. 3 



> or 



1.2. x 1.2. x 8.9.10.11. 11 

 That one shall be white and three black (p = J. 

 g = 3), the chance is 



