108 
Proceedings of the Royal Society 
Then 
and xy = p 1 p 2 , 
therefore, hy means of the theorem proved 
0 - 
y = ViVz + o ( x - ih)> 
>PiP 2 > 
< PiP 2 + 
Todhunter assumes that ^ = 1 
P 2 
and Woolhouse that 
0. 
The above solution contradicts the first three, and agrees with 
the fourth, without introducing more than one unknown quantity. 
But, by means of the theorem referred to, we can find the solution 
when there are n persons involved in the tradition. 
(3.) A 1 says that A 2 says that A 3 says . . . that A n says a 
certain event took place. The probabilities of A 1 , A 2 , . . . A n 
speaking the truth are p v p 2 , . . . p n respectively. Required the 
probability that the event took place. 
Let — 
U = series of statements of A x about A 2 saying &c. , 
x x = which truly reported a statement of A 2 , 
x 2 — ,, ,, A 3 , 
X„ — 
the event. 
Now, 
x n = 
x x x 2 ...x n 
XyC 2 • • • — 1 
= x x x 2 ../x n + ^'(1 - x 1 x 2 ...x n _ 1 ) by the theorem, 
= p 1 p 2 ...p n + 0 (! - PiP2---Pn-i) by the data. 
Cor. 1. Suppose that each always reports truly. Then 
«» = 1 + J x 0 = 1 . 
Cor. 2. Suppose that each always reports truly excepting A n , 
Then 
x n — p n • 
