Unconditional probability I understand the tree diagram but my answer is wrong. 
Urn I contains three red chips and one white chip. Urn II contains two red chips and two white chips. One chip is drawn from each urn and transferred to the other urn. Then a chip is drawn from the first urn.What is the probability that the chip ultimately drawn from urn I is red?

Urn I: 3 red, 1 white.
Urn II: 2 red, 2 white.
There are 4 chips in each urn, and in the first draw is an even exchange of chips (1:1), so we need only worry about how many red chips are in Urn 1 after the exchange ($R$).  The number of red chips can increase by 1, decrease by 1, or remain the same.
$$\mathsf P(R=2) = \frac 3 8,\; \mathsf P(R=3) = \frac 3 8+ \frac 1 8,\; \mathsf P(R=4) = \frac 1 8$$
Now the probability of ultimately drawing a red chip from Urn I is:
$$\begin{align}
\mathsf P(U) & = \mathsf P(U\mid R=2)\mathsf P(R=2)+\mathsf P(U\mid R=3)\mathsf P(R=3)+\mathsf P(U\mid R=4)\mathsf P(R=4)\\[2ex] & = \frac 2 4 \cdot\frac 3 8 + \frac 3 4\cdot(\frac 3 8 +\frac 1 8)+ 1\cdot\frac 1 8 
\\[2ex] & = \frac 3 {16} + \frac 9 {32} +\frac 3 {32} + \frac 1 8
\\[2ex] & = \frac {11}{16}\end{align}$$
These should correspond to numbers on your diagram.  Can you see how they are derived?
