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?
[3/76
Answer:
Fortynineseventytwoths.
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 × = ; in case (2), 1 × ; in case (3), × 1 = ; in case (4), × = ; in case (5), × 1 = ; and in case (6), × = .
Hence the chances of existence for these 6 states, are proportional to ‘12, 3, 18, 2, 9, 4’. Hence their actual values are ‘¼, , , , , , ’.
Hence the chance of the unknown counter being red is the sum of ¼ × 1, × 1, × , × , × , × ; i.e. it is
(36 + 9 + 36 + 4 + 9 + 4)/ 
(9 × 16) 
; which = = .
Q.E.F.