1009 
It is easy to say how many have passed over A or B. Let us 
confine ourselves first of all to A alone. When a particle reaches 
A, the chances that it will lie on the leftside or on the rightside of 
A some time later, will be equally great. Hence the number of 
particles N, that reaches A, is equal to double the number on the 
righthandside of A. 
; EE oO Degas 
ry a= AN VZ] e * dx 
at 
i 
The number of particles that has passed 5 is of course equally 
great. The required number J/, which has neither reached A nor 
B, is however not equal to M—2 N,, for we have counted the 
particles that have passed over the path OAB among the two 
groups V,, as also the particles OBA. (The meaning of this way 
of writing is clear.) The number J, that has travelled the path 
OAB is equally great as the number that has reached a point C 
at a distance 3/ from QO. This is of course again: 
j a ES ax? 
a mmm 
NN es e Erde. 
mt 
31 
Thus we find: 
M—=N—2N,+2N,. 
Now, however, the particles OABA and OBAB have again been 
counted among each of the groups N, ete. Continuing in this way 
we find: 
M= N—2N, +2N,—2N, +... ad enf. . . . (8) 
in which: 
i Go ar” 
a _—— 
Nun cay fl ee ; if, e t dx. 
met 
2m—1)l 
_In order to find the chance P(tdt that a particle for the first 
time passes one of the points A or B between ¢ and ¢-+ di, we 
must differentiate (8) with respect to ¢: 
1 dM 
P (t) dt = — — —d 
Nd 
ee ed ee mete / | 
M= N— AN zr | e t dx — e tdz + f boe ae inf | (9) 
a 
i 31 ‘ 51 
This series is convergent for all finite values of t. 
If we now introduce a new variable y, so that: 
2 
2 ar 
if en 
y p 
