(9) Given the inequality $\frac{x-5}{x+1}<0$ with the solution set $M$, if $x_{0} \in M$, then the probability of $\log _{2}\left(x_{0}+\right.$ 1) $<1$ is ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{1}{5}$
(D) $\frac{2}{5}$
(9) B Hint: $M=\{x \mid-1<x<5\}, \log _{2}(x+1)<1$ has the solution set $\{x \mid-1<x<1\}$, so the probability is $\frac{1}{3}$.