[Math Lair] Pillow-Problems: Problem #23

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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

Answer:

23.

Two-fifths.

Solution:

23.

The original chances, as to states of bag, are

for2W . . . . . ¼;
1W, 1B . . .½;
2B . . . . .¼.
∴ the chances, after adding 2W and 1B, are
for4W, 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.