Problem:
If an infinite number of rods be broken: find the chance that one at least is broken in the middle.
[5/84
Answer:
.6321207 &c.
Solution:
Divide each rod into (n + 1) parts, where n is assumed to be odd, and the n points of division are assumed to be the only points where the rod will break, and to be equally frangible.
The chance of one failure is
;
∴ " " n failures is (
)^{n}
= (1 −
)^{n}
Now, if m =
; then, when n = , m = 0;
∴ the chance that no rod is broke in the middle = (1 − m)^{(1)/(m)}, when m = 0;
i. e. it approaches the limit (1 − 0)
.
And Ans. = 1 − (1 − 0)
.
Now (1 − 0)
= e. Hence if, in the series for e, we call the sum of the odd terms 'a', and, of the even terms 'b'; then e = a + b; and (1 − 0) = a − b = 2a − e.
Q.E.F.
[N.B. What follows here was not thought out.]
Now a = 1 +
+ + &c.
1 = 1
= .5
= .04166666 &c.
= .00138888 &c.
= .00002480 &c.
= .00000027 &c.
∴
a = 1.5430806 & c.
∴ 2
a = 3.0861612 & c.
e = 2.7182818 & c.
∴ (1 − 0)
^{(1)/(10!)} = .3678793 &c.
∴ Ans = 1 − (1 − 0)
^{(1)/(10!)} = .6321207 &c.