Exercise 1.38 An urn contains $1$ green ball, $1$ red ball, $1$ yellow ball and $1$ white ball. I draw $3$ balls with replacement. What is the probability that exactly two balls are of the same color?
So initially I was thinking we'd do $\frac{4}{4}\times\frac{3}{4}\times\frac{2}{4}$ since first draw can be any ball, second must be a different ball, and the third must be one of the initial two, however that's definitely incorrect.
Best Answer
Yes, you've not considered the probability that the first two balls are the same colour and the third different. $+(4/4)(1/4)(3/4)$
Alternatively: count the ways to select two colours for pair and singleton, then arrange them.
$$\dfrac{\binom{4}{1}\binom{3}{1}\binom{3}{2}}{4^3}=\dfrac{4\cdot 3\cdot 2+4\cdot 1\cdot 3}{4^3} =\dfrac{9}{4^2}$$