CHAP. XX.] PROBLEMS ON CAUSES. 335 



The formula (2) may be verified in a large number of cases. 

 As a particular instance, let q = c, we find 



Prob. xy = ab. (3) 



Now the assumption q = c involves, by Definition (Chap. XVI.) 

 the independence of the events B and E. If then B and E are 

 independent, no relation which may exist between A and E can 

 establish a relation between A and B ; wherefore A and B are 

 also independent, as the above equation (3) implies. 



It may readily be shown from (2) that the value of Prob. z, 

 which renders Prob. xy a minimum, is 



Prob z 



If p = <?, this gives 



Prob. z = p ; 



a result, the correctness of which may be shown by the same con- 

 siderations which have been applied to (3). 



PROBLEM V. Given the probabilities of any three events, 

 and the probability of their conjunction ; required the proba- 

 bility of the conjunction of any two of them. 



Suppose the data to be 



Prob. x = p, Prob. y = q, Prob. z = r, Prob. xyz = m, 

 and the quaesitum to be Prob. xy. 



Assuming xyz -s, xy = t, we find as the final logical equa- 

 tion, 



t = xyzs + xyz~s+Q(xy~s + x~s) + - (sum of all other constituents) ; 



whence, finally, 



Prob . xy . 



wherein p = 1 -p, &c. H= p~q + (p + q)r. 



This admits of verification when p = 1, when 0=1, when r = 0, 

 and therefore m = 0, &c. 



Had the condition, Prob. z = r, been omitted, the solution 

 would still have been definite. We should have had 



