An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is

Step-by-step solution

Let W_i and B_i denote the events that the i -th ball drawn is white or black respectively. The urn initially contains 2 white and 2 black balls. Total balls = 4. For the third ball to be black, we consider all possible sequences of the first two draws (D₁, D₂) : 1. W₁, W₂ : P(W₁) = 2 4 = 1 2 . After W₁ , urn has 1W, 2B. P(W₂|W₁) = 1 3 . After W₂ , urn has 0W, 2B. P(B₃|W₁, W₂) = 2 2 = 1 . Probability = 1 2 1 3 1 = 1 6 . 2. W₁, B₂ : P(W₁) = 1 2 . After W₁ , urn has 1W, 2B. P(B₂|W₁) = 2 3 . After B₂ , urn has 1W, 3B