# Pillow-Problems: Problem #45

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Problem #44 | Problem #47
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

## Problem:

45.

If an infinite number of rods be broken: find the chance that one at least is broken in the middle.

[5/84

45.

.6321207 &c.

## Solution:

45.

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 − 1)/ (n)
;

∴ " " n failures is (
 (n − 1)/ (n)
)n

= (1 −
 (1)/ (n)
)n

Now, if m =
 (1)/ (n)
; then, when n =
 (1)/ (0)
, 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)
 (1)/ (0)
.

And Ans. = 1 − (1 − 0)
 (1)/ (0)
.

Now (1 − 0)
 (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)
 (1)/ (0)
= ab = 2ae.

Q.E.F.

[N.B. What follows here was not thought out.]

Now a = 1 +
 (1)/ (2!)
+
 (1)/ (4!)
+ &c.

1 = 1
 (1)/ (2!)
= .5
 (1)/ (4!)
= .04166666 &c.
 (1)/ (6!)
= .00138888 &c.
 (1)/ (8!)
= .00002480 &c.
 (1)/ (10!)
= .00000027 &c.
a = 1.5430806 & c.
∴ 2a = 3.0861612 & c.
e = 2.7182818 & c.
∴ (1 − 0)
 (1)/ (10!)
= .3678793 &c. ∴ Ans = 1 − (1 − 0)
 (1)/ (10!)
= .6321207 &c.