A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colours without taking them out of the bag.
One is black, and the other white.
We know that, if a bag contained 3 counters, 2 being black and one white, the chance of drawing a black one would be
; and that any other state of things would not give this chance.
(2)/ (3)
Now the chances, that the given bag contains (α) BB, (β) BW, (γ) WW, are respectively ¼, ½, ¼.
Add a black counter.
Then the chances, that it contains (α) BBB, (β) BWB, (γ) WWB, are, as before, ¼, ½, ¼.
Hence the chance, of now drawing the black one,
Hence the bag now contains BBW (since any other state of things would not give this chance).
hence, before the black counter was added, it contained BW, i.e. one black counter and one white.
Q.E.F.
+ ¼ · (2)/ (3)
= (1)/ (3)
.(2)/ (3)