An urn contains four white balls and two red balls. A ball is drawn at random and is replaced in the urn each time. What is the probability that after two successive draws, both the balls drawn are white.
Clearly there are two ways this could go.
Case 1. First white ball is drawn $\rightarrow$ It is replaced in the urn by another white ball $\rightarrow$ a second white ball is drawn
Case2. First white ball is drawn $\rightarrow$ it is replaced in the urn by a red ball $\rightarrow$ second white ball is drawn
So the final probability should be $\frac{4}{6}\frac{4}{6}+\frac{4}{6}\frac{3}{6}=\frac{7}{9}$. But this does not match the answer given in the book.
Best Answer
As we replace the balls, we can say that the trials are independent. So, it boils down to find the probability $P (WW) $ denoting white balls in two successive draws. Thus, we get, $$P (WW) =P (W)P (W) =(\frac {4}{6})^2 =\frac {4}{9}$$
The mistake in your working was that you considered the second case needlessly. We do not replace any coloured ball, we replace the one that is taken out. That is to be remembered. Hope it helps.