7. A bag contains 6 red balls and 8 white balls. If 5 balls are randomly taken and placed in box $A$, and the remaining 9 balls are placed in box $B$, then the probability that the sum of the number of white balls in box $A$ and the number of red balls in box $B$ is not a prime number is $\qquad$. (Answer with a number)
Given the number of white balls in box $A$ is $m$, and the number of red balls in box $B$ is $n$, then 
\[
\left\{\begin{array}{l}
0 \leqslant m \leqslant 5, \\ 
0 \leqslant n \leqslant 6, \\ 
m+9-n=8, \\ 
5-m+n=6,
\end{array}\right.
\]
Thus, the combinations where $m+n$ is a prime number are $(1,2),(2,3),(3,4),(5,6)$.
Let $E=$ “the sum of the number of white balls in box $A$ and the number of red balls in box $B$ is not a prime number”, then 
\[
P(\bar{E})=\frac{C_{8}^{1} \cdot C_{6}^{4}+C_{8}^{2} \cdot C_{6}^{3}+C_{8}^{3} \cdot C_{6}^{2}+C_{8}^{5}}{C_{14}^{5}}=\frac{788}{1001},
\]
so 
\[
P(E)=1-P(\bar{E})=\frac{213}{1001}.
\]