Pillow-Problems: Problem #38

Problem #32 | Problem #39
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

There are 3 bags, 'A', 'B', and 'C'. 'A' contains 3 red counters, 'B' two red and one white, 'C' one red and 2 white. Two bags are taken at random, and a counter drawn from each: both prove to be red. The counters are replaced, and the experiment is repeated with the same two bags: one proves to be red. What is the chance of the other being red?

Answer:

Fortynine-seventytwoths.

Solution:

Taking, in order, the bag from which this unknown counter is drawn, the bag from which a red one was twice drawn, and the remaining bag, we see that there are six possible arrangements of ‘A’, ‘B’, and ‘C’: viz.—

(1) ABC,	(4) BCA,
(2) ACB,	(5) CAB,
(3) BAC,	(6) CBA.

Now the chance of the observed event is, in case (1), 1 ×

(4)/

(9)

=

(4)/

(9)

; in case (2), 1 ×

(1)/

(9)

; in case (3),

(2)/

(3)

× 1 =

(2)/

(3)

; in case (4),

(2)/

(3)

×

(1)/

(9)

=

(2)/

(27)

; in case (5),

(1)/

(3)

× 1 =

(1)/

(3)

; and in case (6),

(1)/

(3)

×

(4)/

(9)

=

(4)/

(27)

.

Hence the chances of existence for these 6 states, are proportional to ‘12, 3, 18, 2, 9, 4’. Hence their actual values are ‘¼,

(1)/

(16)

,

(3)/

(8)

,

(3)/

(8)

,

(1)/

(24)

,

(3)/

(16)

,

(1)/

(12)

’.

Hence the chance of the unknown counter being red is the sum of ¼ × 1,

(1)/

(16)

× 1,

(3)/

(8)

×

(2)/

(3)

,

(1)/

(24)

×

(2)/

(3)

,

(3)/

(16)

×

(1)/

(3)

,

(1)/

(12)

×

(1)/

(3)

; i.e. it is

(36 + 9 + 36 + 4 + 9 + 4)/

(9 × 16)

; which =

(98)/

(9 × 16)

=

(49)/

(72)

.

Q.E.F.