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?
[25/9/87
Answer:
Twofifths.
Solution:
The original chances, as to states of bag, are
for  2W . . . . .  ¼;

 1W, 1B . . .  ½;

 2B . . . . .  ¼.

∴ the chances, after adding 2W and 1B, are
for  4W, 1B . . . .  ¼;

 3W, 2B . . . .  ½;

 2W, 3B . . . .  ¼.

Now the chances, which these give to the observed event, drawing 2W and 1B, are
,
,
.
∴ the chances, after this event, are proportional to
, , i.e. to 2, 4, 1. Hence they are ,
,
.
Hence the chances, as to states, now are
for  2W . . . . .  ;

 1W, 1B. . . .  ;

 2B . . . . .  . 
∴ the chances, after adding 1W, are
for  3W . . . . .  ;

 2W, 1B. . . .  ;

 1W, 2B. . . .  . 
Now the chances, which these give to the observed event, of drawing 1W, are 1, , .
∴ the chances, after this event, are proportional to , , ; i.e. to 6, 8, 1. Hence they are , , .
Hence the chance, that the bag now contains 2 white, is ; i.e. .
Q.E.F.