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?
 A: Here is a tree diagram for the scenario.

A reminder on how to use the tree diagram, on each branch the probability of traveling along that branch from the previous branching point is written.  For example, to travel along the topmost branch on the left corresponds to the event of pulling a red from the first urn and a white from the second, which occurs with probability $\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}$.  To arrive at a particular leaf corresponds to having traveled along the branches to get there, and so occurs with probability equal to the product of the probabilities associated with each branch.  For example, the topmost leaf is $\frac{3}{8}\cdot\frac{1}{2}=\frac{3}{16}$.
The event we are curious about is "pull a red from urn 1 at the end", and the corresponding leaves for that event are put in blocks.  Adding these together gives the final answer $\frac{3}{16}+\frac{9}{32}+\frac{3}{32}+\frac{1}{8}=\frac{11}{16}$.
A: Hint:
Urn I should end up with the following chips given the color of chip being taken from I and II and transferred accordingly.
$$
\begin{array}{c|c|}
Grid & \text{Red(I)} & \text{White(I)} \\
\hline
\text{Red(II)} & \text{3r+1w}& \text{4r} \\
\text{White(II)} & \text{2r+2w} & \text{3r+1w} 
\end{array}
$$
A: 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?
