















AND ITS APPLICATION TO A PROBLEM IN PROBABILITIES. 545 
Then the series written perpendicularly is 
m—1" x1 3m—1" x1 3m—1" x1 3,m—1" x1 3m—1" x1 
m—2"x8 | 3m—2" x7 3,m—2" x 6 zym—2? x5 3,m—2" x 4 
m—8" x27 = 3,m—3!" x19 = 3,m—3" x12 = 3m—3?x6 = 3m—3" x1 
m—4" x64  %m—4"?x37 %m—4?x18 3,m—4"x6 
m—5"x125 3m—5? x61 3m—5" x20 = 35m—5"’ x6 
&e. &e. &e. &e. 
The value of the different sigmas is easily found by the method of finite dif- 
ferences. 
Generally, since the differences of 1’, 2%, 3°, &c., always vanish in the g+1'* 
line and after the g™ term of it, the general expression is 
By41m— 1” + dy 2y41mM—2QW. .. . dy 3q41m—q"; 
d,, d,, d,, &¢., signifying the 1st, 2d, 3d, &c., terms of the g+1' row of differences. 
* This summation may be applied to find the probability in the case now under 
consideration, for it expresses the 7+ (+ ‘y, &c., of the preceding case. Applying it 
as we did the value of 2+ 8 +7, &c., there found, we shall find the probability of 
a white ball at the p+q+1™ trial to be 






3,1 dy Sym — 2°")... dy Bou m—qrn K) 
m (3gy1m— 1"? + dy 2y41m—2Q?. 2. . dy Bq41m—Q” ( 
If m be infinite, the expression becomes 
(l+d,+.... dy) «Bg mP*! _3y41,mP*? 
m (L+d,.. ~~ dy)-3y41mP — M3q41m” 
But if z be a quantity varying between the limits 0, 2, 
‘0 
z sata ee _ pti gprt? 
aim a [rode Pa eee 
And by continuation 
Zyyim?t! ptl.p+2.... p+qtl mn pt+l cL) 
MIyi1mP p+2.pt+3....ptqt+2 ptpt2 ~ —"— © 
We have thus found the probability in every case of the problem; the 2d and 
4th at H, for the result, being independent of m, must be true for an infinite as 
-_-well as for a finite number. The Ist case is solved at K, and the 3d at L. 
3 VOL. XX. PART IV. 7H 
