# Probability problem related to drawing balls from an urn

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.

• This is not clear. In these sort of problems "replace" generally means "put back". That is, you take out a ball, look at it, and return it to the urn. You appear to mean something else. If you are adding a new ball from some supply we need to know the color make up of the supply. – lulu Jan 18 '17 at 11:34
• The wording just means that each time you draw, you are drawing from (4 white + 2 red) balls, so can you figure it out now ? – true blue anil Jan 18 '17 at 11:37
• @lulu Yes that does it! – Miz Jan 18 '17 at 11:38

Replaced in these type of questions mean we put back same type or colour of item. So we replace white ball after first pick.

So your second case is unnecessary.

Probability = $\frac46 \times \frac46 = \frac49$

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.