6. In an accident, a maternity hospital lost all the labels of 8 babies born on that day. Without any markings, the probability that exactly 4 families take the right babies is $\qquad$ .
6. $\frac{1}{64}$ Detailed Explanation: Applying the principle of inclusion-exclusion, the number of ways for all four families to be completely mismatched is $A_{4}^{4}-4 A_{3}^{3}+6 A_{2}^{2}-4 A_{1}^{1}+1 A_{0}^{0}=9$. Therefore, the probability is $\frac{C_{8}^{4} \cdot 9}{A_{8}^{8}}=\frac{1}{64}$.