9. Given that $A B C D$ is a tetrahedron, and each edge is colored red or blue with equal probability, the probability that point $A$ can reach point $B$ through the red edges of the tetrahedron $A B C D$ is $\qquad$ .
When $A B$ is colored red, the probability is $p_{1}=\frac{1}{2}$, which satisfies the condition;
When $A B$ is colored blue, the probability is $\frac{1}{2}$. As shown in the figure, consider the coloring of $A C, A D$:
$A C, A D$ are both colored red, the probability is $\frac{1}{4}$, in this case, either $B C$ or $B D$ needs to be colored red.
The probability is $p_{2}=\frac{1}{8}\left(1-\frac{1}{4}\right)=\frac{3}{32}$;
$A C, A D$ are both colored blue, which does not satisfy the condition;
$A C, A D$ are colored one red and one blue, the probability is $\frac{1}{2}$. Without loss of generality, assume $A C$ is red and $A D$ is blue, in this case, either $B C$ is red, or both $B D$ and $C D$ are red, the probability is $p_{3}=\frac{1}{4}\left[1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\right]=\frac{5}{32}$. In summary, let $E=$ “point $A$ can reach point $B$ through the red edges of tetrahedron $A B C D$”, then $P(E)=p_{1}+p_{2}+p_{3}=\frac{1}{2}+\frac{3}{32}+\frac{5}{32}=\frac{3}{4}$.