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:

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?

Answer:

Two-fifths.

Solution:

The original chances, as to states of bag, are

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.