7. A bag contains 6 red balls and 8 white balls. Any five balls are placed in box $A$, and the remaining nine balls are placed in box $B$. 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).
7. $\frac{213}{1001}$.

Let the number of white balls in box $A$ be $n (0 \leqslant n \leqslant 5)$, then the number of red balls in box $A$ is $5-n$; the number of white balls in box $B$ is $8-n$, and the number of red balls is $n+1$. Therefore, the sum of the number of white balls in box $A$ and the number of red balls in box $B$ is $2 n+1$, and $1 \leqslant 2 n+1 \leqslant 11$.

Notice that $2 n+1$ is an odd number, and the non-prime odd numbers are only 1 or 9, i.e., $n=0$ or 4.
Thus, the required probability $P=\frac{\mathrm{C}_{6}^{5}+\mathrm{C}_{8}^{4} \mathrm{C}_{6}^{1}}{\mathrm{C}_{14}^{5}}=\frac{213}{1001}$.