Answer
$ \frac{11}{16}$
Work Step by Step
Define the events
$A_{1}$ = "Red chip transferred from urn 1. Red chip transferred from urn 2."
$A_{2}$ = "Red chip transferred from urn 1. White chip transferred from urn 2."
$A_{3}$ = "White chip transferred from urn 1. White chip transferred from urn 2."
$A_{3}$ = "White chip transferred from urn 1. Red chip transferred from urn 2."
$B$ = "After the transfer, the chip drawn from urn $1$ is red."
The probabilities of $A_{i}$ are as follows:
$P(A_{1})= \displaystyle \frac{3}{4}\cdot\frac{2}{4} =\frac{3}{8} $
(chosen: one of three reds from urn 1, one of 2 reds from urn 2)
$P(A_{2})=\displaystyle \frac{3}{4}\cdot\frac{2}{4}=\frac{3}{8}$
(chosen: one of three reds from urn 1, one of 2 whites from urn 2)
$P(A_{3})=\displaystyle \frac{1}{4}\cdot\frac{2}{4}=\frac{1}{8}$
(chosen: one of one whites from urn 1, one of 2 whites from urn 2)
$P(A_{4})=\displaystyle \frac{1}{4}\cdot\frac{2}{4}=\frac{1}{8}$
(chosen: one of one whites from urn 1, one of 2 reds from urn 2)
The conditional probabilities $P(B|A_{i})$ are as follows:
$P(B|A_{1}) =P($Urn 1 contains $3$R and $1$W. A red is drawn.$)=\displaystyle \frac{3}{4}$
$P(B|A_{2}) =P($Urn 1 contains $2$R and $2$W. A red is drawn.$)=\displaystyle \frac{2}{4}=\frac{1}{2}$
$P(B|A_{3}) =P($Urn 1 contains $3$R and $1$W. A red is drawn.$)=\displaystyle \frac{3}{4}$
$P(B|A_{4}) =P($Urn 1 contains $4$R and $0$W. A red is drawn.$)=\displaystyle \frac{4}{4}=1$
We can now apply the theorem
$$\begin{align*}
P(B)&=\displaystyle \sum_{i=1}^{n}P(B|A_{i})P(A_{i}) & & \\\\
&= \displaystyle \frac{3}{4}\cdot\frac{3}{8} + \frac{1}{2}\cdot\frac{3}{8} + \frac{3}{4}\cdot\frac{1}{8} + 1\cdot\frac{1}{8} \\\\
& =\displaystyle \frac{11}{16} \end{align*}$$
