[Math] Probability: Third Draw from an urn having black and white balls

Question

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
with another ball of same colour. The process is repeated. Find the
probability that the third ball drawn is black.

I am getting the final answer as $5/8$ from my tree diagram. But in my book the answer is $23/30$. I don't know where I am going wrong. Please help.

Answer 1

The question means to say that

If you draw a white ball, then don't put back. If you draw black ball, then put two black balls back.

Or put another way,

If it is white, it is not replaced (put back) into the urn. Otherwise it is replaced with alongside another ball of same color. The process is repeated. Find the probability that the third ball drawn is black.

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Let $B_i,W_i$ be the events that you drew a white or black ball in the $i$th draw. Then,
\begin{align*} P(B_3) &=P(B_2B_3)+P(W_2B_3)\\ &=P(B_1B_2B_3)+P(W_1B_2B_3)+P(B_1W_2B_3)+P(W_1W_2B_3)\\ &=P(B_3|B_2B_1)P(B_2|B_1)P(B_1)+P(B_3|B_2W_1)P(B_2|W_1)P(W_1)\\ &\qquad+P(B_3|W_2B_1)P(W_2|B_1)P(B_1)+P(B_3|W_2W_1)P(W_2|W_1)P(W_1)\\ &=\frac46\frac35\frac24+\frac34\frac23\frac24+\frac34\frac25\frac24+\frac22\frac13\frac24\\ &=\frac{23}{30}. \end{align*}