ON INVERSE PROBABILITIES. 57 



is not known, is a black- ball. A ball is drawn and 

 replaced ; and this process is repeated, but whether out 

 of the same urn as before is not known. Both draw- 

 ings give a white ball : what is the probability of the 

 several cases from which this result might have hap- 

 pened ? 



Since the black ball is as likely to be in one as in 

 the other, the antecedent state of things is (so far as a 

 single drawing is concerned,,) the same as if there were 

 four urns, as follows : 



I. (3 white) 



II. (3 white, 1 black) 

 IV. (2 white, 1 black). 



III. (2 white) 



There are 16 possible cases, PP [2, 4,] numbered in 

 the first columns following, described in the second, 

 and having the probability which each would give 

 to the observed event (both drawings white) registered 

 in the third, together with the numerator, when all the 

 fractions are reduced to a common denominator 1 44. 



That is to say, if (case 8), II. and IV. were the urns 

 of the first and second drawings, the chance of the 

 observed event is \ or T ^. But, it must be remem- 

 bered, that we do not suppose the black ball may have 

 been removed from one urn into the other before the 

 second drawing takes place. Most of the preceding 

 cases are, therefore, to be rejected; in fact, I. can 

 combine with nothing but I. or IV., and II. with 

 nothing but II. or III. Reject, therefore, cases 2, 3, 

 5, 8, 9j 12, 14, 15, and the sum of the numerators in 

 the rest is 841. Hence the probability (for instance,) 



