Probability Paradox: One White and One Black Marble

[soroban]

I saw this "proof" many years ago.
.I thought you might enjoy it.A bag contains two marbles.
Either can be Black or White.
Determine the colors of the marbles.

Answer: one Black marble and one White marble.$$\text{Proof}$$

$$\text{There are three equally likely situtations.}$$
. . $$\text{The bag contains: }\:BB,\,BW,\,WW.$$

$$\text{Add one White marble to the bag.}$$$$\text{Then we have:}$$

$$[1]\;P(BBW) \,=\,\tfrac{1}{3}$$
. . $$P(W\,|\,BBW) \:=\:\left(\tfrac{1}{3} \right)\left(\tfrac{1}{3} \right) \:=\:\tfrac{1}{9}$$

$$[2]\;P(BWW) \,=\,\tfrac{1}{3}$$
. . $$P(W\,|\,BWW) \:=\:\left(\tfrac{1}{3}\right)\left(\tfrac{2}{3} \right) \:=\:\tfrac{2}{9}$$

$$[3]\;P(WWW) \,=\,\tfrac{1}{3}$$
. . $$P(W\,|\,WWW) \:=\:\left(\tfrac{1}{3}\right)\left(\tfrac{3}{3} \right) \:=\:\tfrac{3}{9}$$

$$\text{Hence: }\:P(W) \:=\:\tfrac{1}{9}\,+\,\tfrac{2}{9}\,+\,\tfrac{3}{9} \:=\:\tfrac{6}{9}\:=\:\tfrac{2}{3}$$$$\text{The probability of drawing a White ball is }\tfrac{2}{3}.$$
. . $$\text{The bag }must\text{ contain 2 White balls and 1 Black ball.}$$$$\text{Therefore, before the White ball was added,}$$
. . $$\text{the bag had one White ball and one Black ball.}$$

$$\text{Q.E.D.}$$

 

[DavidCampen]

Are we supposed to point out the fallacy?