596 
NOTES AND REFERENCES. 
The formulae show that p 2 is real; and then further, taking for p its positive 
value, it at once appears that we have u, X, pb no one of them negative or greater 
than 1, viz. the values are such as these quantities, as probabilities, ought each of 
them to have: and we have thus a real solution. 
Boole in 1 after remarking that the quadratic equation in u may be written 
in the form 
(1 -ap'-u) (l-ßq'-u) = a , , 
1 — u ^ 
(p' = 1 —p, &c.) 
says that this is certainly erroneous; for in the particular case p = 1, ^=0 it gives 
u = 1 or u = a (1 — /3), whereas the value should be u = a. But observe that p = l, q = 0, 
give q — ap, — — a, a negative value, so that the solution does not apply. If we further 
examine the meaning, A is a cause such that if it act then (p = 1) the event is 
sure to happen; and I? is a cause (?) such that if it act then (q = 0) the event is 
sure not to happen; this is self-contradictory unless we make the new assumption 
that the causes A and B cannot both act. It is remarkable that even in this case 
my solution gives the plausible result u = a(l—/3), viz. the probability of the event is 
the product of the probabilities of A acting, and B not acting. 
In further illustration, and at the same time to examine Boole’s solution, I 
write as follows: 
Wilbraham. Boole. Cayley. 
ABE 
X y st 
aß (1 — A!p!) 
ABE’ 
f 
x y s't' 
a ß X'pf 
ABE 
V 
x'y s't 
a ßp. 
ABE’ 
V 
Ay s't' 
a ßp! 
AB’E 
t 
x y's 
a ß'X 
AB’E’ 
C 
x y s't' 
a ß'X' 
A’B’E 
0 
0 
0 
A'B’E' 
or’ 
x'y's't' 
a'ß' 
where in the first column the accent denotes negation: ABE means that the events 
A, B, E all happen, ABE' that A and B each happen, E' does not happen, and 
so for the other symbols. And in like manner in the third and fourth columns, where 
the unaccented letters denote probabilities, an accented letter is the probability of the 
contrary event, x' = 1 — x, &c. 
By hypothesis E cannot happen unless either A ox B happen, that is Prob. 
A'B'E = 0, or writing A'B'E for the probability (and so in other cases) say A'B'E = 0. 
And I then (with Wilbraham) denote the probabilities of the other seven combinations 
of events by f, £', rj, r[, £, £' and A ; and (as before) the required probability of the 
event E by u. 
The data of the Problem are 1 = 1, A = a, B — ¡3, AE = ap, BE = ¡3q, and we have 
thence to find E = u, where on the left-hand side of the first equation 1 means 
ABE 4- ABE' + &c. = f + 77 4- mi + £ + + a, and similarly A means 
ABE+ABE' + AB'E + AB'E', = £ + f 4- £+ & c .;