# Pillow-Problems: Problem #38

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

## Problem:

38.

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?

[3/76

38.

Fortynine-seventytwoths.

## Solution:

38.

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.