NOTES AND REFERENCES. 
595 
produce the event E, but that the event E cannot happen unless at least one of 
them act efficiently, viz. it cannot happen in consequence of the conjoint separately 
inefficient action of the two causes. On these assumptions it appears to me that my 
solution, as completed by Dedekind, is correct. This would not preclude the correct 
ness of Boole’s solution, if according to what precedes we consider it as the solution 
of a different question: but I am unable to understand it. 
I resume my own solution, completing it according to Dedekind. I write with him 
u instead of p for the required probability of the event E; the equations of the 
text thus are 
p = X + (1 — X) piß, q = p + (1 — p)Xa; u = Xa + pß — Xpaß, 
and we thence deduce 
u — ßq = (1 — /3) Xa, u — ap = (1 — a) pß; 
and then eliminating X, p, we find 
_u — ßq u — ap (u — ßq) (u — ap) 
U = 1-J + T^T “ (T - /37(1 - a) ’ 
or as this equation may be written 
u 2 — u (1 — aß + ap + ßq) + (1 — ß) ap + (1 —a)ßq + aßpq = 0 ; 
say we have 
where 
u = \ (1 - aß + ap + ßq — p), 
p 2 = (1 — aß + ap + ßq) 2 — 4 (1 — ß) ap — 4 (1 — a) ßq — 4<aßpq, 
= (1 —2a + aß + ap — ßq) 2 + 4a (1 — a) (1 — ß) (1 — p), 
= (1 - 2/3 + aß - ap + ßq) 2 + 4/3 (1 - ß) (1 - a) (1 - q), 
= (1 - aß + ap - ßq) 2 - 4a (1 - ß) (p - ßq), 
= (1 — aß — ap + ßq) 2 — 4/3 (1 — a) (q — ap), 
and hence also 
(1 - aß +■ ap - ßq - p) 
(l-ß)'X 
k(l-aß- ap + ßq~ p) 
(!-«)£ 
Here p, q, a, /3, as probabilities, are none of them negative or greater than 1; p is 
the probability that, A acting, E will happen; and ¡3q is the probability that B will 
act and E happen. But if A act, then even if B does not act, E may happen, or 
B may act and E happen, that is p is greater than or at least equal to /3q, say p — ¡3q 
is not negative. And similarly q — ap is not negative. We thus have as conditions 
of a possible experience, p — (3q and q — ap neither of them negative. 
75—2