# Pillow-Problems: Problem #23

Math Lair Home > Source Material > Pillow-Problems > Problem #23
Problem #22 | Problem #24
Note that the source contains a minor typo in the solution, which I have preserved below. "
 (3)/ ()
,
 (3)/ (5)
,
 (3)/ (10)
" should read "
 (3)/ (5)
,
 (3)/ (5)
,
 (3)/ (10)
."
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

## Problem:

23.

A bag contains 2 counters, each of which is known to be black or white. 2 white and a black are put in, and 2 white and a black drawn out. Then a white is put in, and a white drawn out. What is the chance that it now contains 2 white?

[25/9/87

23.

Two-fifths.

## Solution:

23.

The original chances, as to states of bag, are

 for 2W . . . . . ¼; 1W, 1B . . . ½; 2B . . . . . ¼.
∴ the chances, after adding 2W and 1B, are
 for 4W, 1B . . . . ¼; 3W, 2B . . . . ½; 2W, 3B . . . . ¼.

Now the chances, which these give to the observed event, drawing 2W and 1B, are
 (3)/ ()
,
 (3)/ (5)
,
 (3)/ (10)
.

∴ the chances, after this event, are proportional to
 (3)/ (20)
,
 (3)/ (10)
,
 (3)/ (40)
i.e. to 2, 4, 1. Hence they are
 (2)/ (7)
,
 (4)/ (7)
,
 (1)/ (7)
.

Hence the chances, as to states, now are
for 2W . . . . .
 (2)/ (7)
;
1W, 1B. . . .
 (4)/ (7)
;
2B . . . . .
 (1)/ (7)
.

∴ the chances, after adding 1W, are
for 3W . . . . .
 (2)/ (7)
;
2W, 1B. . . .
 (4)/ (7)
;
1W, 2B. . . .
 (1)/ (7)
.

Now the chances, which these give to the observed event, of drawing 1W, are 1,
 (2)/ (3)
,
 (1)/ (3)
.

∴ the chances, after this event, are proportional to
 (2)/ (7)
,
 (8)/ (21)
,
 (1)/ (21)
; i.e. to 6, 8, 1. Hence they are
 (6)/ (15)
,
 (8)/ (15)
,
 (1)/ (15)
.

Hence the chance, that the bag now contains 2 white, is
 (6)/ (15)
; i.e.
 (2)/ (5)
.

Q.E.F.