326 PROBLEMS ON CAUSES. [CHAP. XX. 



Similarly, 



< - 



And thus the conditions by which the general solution was 

 limited are confirmed. 



Again, let p l = 1, p 2 = 1. This is to suppose that when either 

 of the causes A 19 A z is present, the event E will occur. We have 

 then a = CD b = c Z9 a = 1, b' 1, c' = Ci -f c 2 , and substituting in 

 (13) we get 



C! - C a - I) 2 + 4 (d C z - C t - C 2 )| 



-2 



= d + c 2 - CiCjj on reduction 



Now this is the known expression for the probability that one 

 cause at least will be present, which, under the circumstances, is 

 evidently the probability of the event E. 



Finally, let it be supposed that c x and c 2 are very small, so 

 that their product may be neglected ; then the expression for u 

 reduces to c l p l + C 2 p 2 . Now the smaller the probability of each 

 cause, the smaller, in a much higher degree, is the probability of 

 a conjunction of causes. Ultimately, therefore, such reduction 

 continuing, the probability of the event E becomes the same as 

 if the causes were mutually exclusive. 



I have dwelt at greater length upon this solution, because it 

 serves in some respect as a model for those which follow, some of 

 which, being of a more complex character, might, without such 

 preparation, appear difficult. 



5. PROBLEM II. In place of the supposition adopted in the 

 previous problem, that the event E cannot happen when both the 

 causes A 19 A z are absent, let it be assumed that the causes A 19 A 2 

 cannot both be absent, and let the other circumstances remain as 

 before. Required, then, the probability of the event E. 



Here, in place of the equation (2) of the previous solution, we 

 have the equation 



(l-*)(l-y)-0. 



The developed logical expression of z is found to be 



