288 ELEMENTARY ILLUSTRATIONS. [CHAP. XVIII. 



We have then, 



Prob. {x + (1 - x) yz} = p, 

 to find Prob. x. 



Let x + (1 - x) yz = s 9 



then eliminating yz as a single symbol, we get, 



x(l-s) = 0. 

 Hence 



whence, proceeding according to the rule, we have 



Prob. x = cp, (1) 



where c is the probability that if x occurs, or alone fails, the 

 former of the two alternatives is the one that will happen. The 

 limits of the solution are evidently and p. 



This solution appears to give us no information beyond what 

 unassisted good sense would have conveyed. It is, however, all 

 that the single datum here assumed really warrants us in infer- 

 ring. We shall in the next solution see how an addition to our 

 data restricts within narrower limits the final solution, 



SOLUTION OF THE SECOND CASE. 



Here we assume as our data the equations 

 Prob. {x + (1 - x) yz} = p, 



Prob. {y + (l-y)xz) = q. 

 Let us write 



from the first of which we have, by (VIII. 7), 



{x+(\-x)yz} (l-s) + s{l-z-(l-x)yz} = 0, 

 or (a; + xyz) ~s + sx (1 - yz) = ; 



provided that for simplicity we write x for 1 - a?, y for 1 - y, and 

 so on. Now, writing for 1 - yz its value in constituents, we 



have 



(x + xyz) s + sx (yz + yz + yz) = 0, 



an equation consisting solely of positive terms. 



