foundations in probability

The first ball is red and the last ball is green. The probabiliy that the secod ball to be red is 
\[\frac{2}{3}\]
and the probability that the second ball to be green is $\frac{1}{3}$. Let $f(n)$ be the probability of getting $n$ red balls. Then 
So \[f(1)=\frac{1}{3}, f(2)=\frac{2}{3}\]
and zero otherwise. The conditional expectation of the number of red balls is
\[1\times f(1)+2\times f(2)=1\times \frac{1}{3}+2\times \frac{2}{3}=\frac{5}{3}.\]