Theory of Probabilities. 35 9 



them equal to unity), i. e. A always acts efficiently. The third 

 equation then gives g=/A + (l — ij)ol } or /J>=—, — , and we then 

 have q — u m 



that is, 



w= a /3' + /3?; 



that is, the probabihty of E is the probability of AB' plus the 

 probability of BE. 



Since q <£ a, we cannot have q = 0) in fact, the two equations 

 jb = 1, #=0 fjL e. if A, then E always; if B, then E never) would 

 imply that A and B could not happen together ; which is con- 

 trary to the hypothesis that they are independent causes. 



Next for Prof. Boole's solution : p = 1, the conditions of applica- 

 bility are satisfied if only + j3q <fcct.; or, what is the same thing, 

 a' <£ j3q' ; viz. A' may happen with BE' or with BE, but BE' can 

 only happen with A', that is, prob. A' <£ prob. BE'. 



The equation in u written under the form 



— (u — l)(u— a)(u— b) 



+ ( u -f){u-ff)(u-h)=0, 



when j9 = l, and therefore /=1, gives immediately (since we 

 cannot have u = l) 



(u—a)(u — b) = (u—g)(u—h); 

 that is, 



u{g + h — a — b)=gh — ab, 

 and therefore 



_ (* + !3q)(ff + l3q)-*13q _ «ff + £ g ( ff + gg) 



U ~~ P + Pq $' + Pq 



or finally, 





In the present case, if only 1 <£a + /3, g = 0is consistent with 

 the condition of applicability ; and it gives, as it ought to do, 

 u = u. 



But q retaining its general value, we have 



s . x "■ — ^ n 



#=0, t = g0 (that is, 5 = 1), -j = 



^ @+Pq 



_ fiq' t _ u—ol 

 y ~~V> 7~~JgJ ; 



