Obtain the mean of X the number of red balls drawn

Question

Urn 1 contains 2R balls and 1W, Urn 2 contains 2W balls and 1R. using a fair coin of head to choose Urn 1 and otherwise 2. Two balls are drawn randomly without replacement from the chosen Urn. X is the number of Red balls. we need the mean of X.

I started by Finding the probability of obtaining all the cases

$$ \begin{array}{c|lcr} n & \text{URN1} & \text{URN2} & \text{P(n|Urn1)+P(n|Urn2)}\\ \hline 2R & 1/2\times2/3\times1/2 & 1/2\times1/3\times0 & 1/6 \\ 1R & 1/2\times2/3\times1/2 & 1/2\times1/3\times1 & 1/3\\ 0R & 1/2\times1/3\times0 & 1/2\times2/3\times1/2 & 1/6 \end{array} $$

the result by the book is 1 while by me after multiplying $xp(x)$ is $0.5$

Answer 1

You only took into account one of two orders of drawing one red and one white ball (for each urn).

Answer 2

See the answer of Joriki for what went wrong.

Actually this can be solved more directly on base of symmetry and linearity of expectation.

Let $Y$ denote the number of white balls drawn so that $X+Y=2$ and consequently $$\mathbb EX+\mathbb EY=\mathbb E(X+Y)=2$$

Then realize that $X$ and $Y$ have the same distribution so that $$\mathbb EX=\mathbb EY$$