# Probability: Third Draw from an urn having black and white balls

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.

• There must be an arithmetical error, $85/72$ is not possible, it is larger than $1$. – André Nicolas Mar 13 '16 at 8:25
• How can you get a final answer that is larger than $1$? Your fractions simplify and add up to $$\frac16 + \frac29 + \frac19 + \frac 18 = \frac58$$ – Arthur Mar 13 '16 at 8:28
• Can you check the wording? A common problem is "it is replaced, together with another ball of the same colour." – André Nicolas Mar 13 '16 at 8:28

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.

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*}

• But the question states "If it is white, it is not replaced into the urn. Otherwise it is replaced with another ball of same colour." – true blue anil Mar 13 '16 at 8:57
• Yeah, that's another way to put what I said. Or, I just rephrased the question. In any case, Whatever the problem meant is what I did so that the answer matches. I'm not saying that the wording is clear. I am saying that what I did is what the problem meant to say. – Em. Mar 13 '16 at 8:59
• Maybe I'm missing something, but for $BBB$, there are $2$ balls out of $4$ at start, and their number doesn't change according to the given condition, so how do you get $4/6 \times 3/5 \times 2/4$ and not just $(2/4)^3$ ? – true blue anil Mar 13 '16 at 9:12
• Oh, hahahaha, I meant if draw white, then don't put back. If you draw black, then put two black back. Sorry, I thought that's what I wrote. – Em. Mar 13 '16 at 9:18
• So if we want the given answer, we need to change the question ! So be it :) – true blue anil Mar 13 '16 at 9:21