14$\cdot$9 A rope is cut into two pieces at a random point. The probability that the longer piece is at least $x$ times the length of the shorter piece is
(A) $\frac{1}{2}$.
(B) $\frac{2}{x}$.
(C) $\frac{1}{x+1}$.
(D) $\frac{1}{x}$.
(E) $\frac{2}{x+1}$.
(23rd American High School Mathematics Examination, 1972)
［Solution］Assume $AB$ is this rope (as shown in the figure), and $P$ is a point on $AB$ such that
$$
AP: PB = 1: x \text{.}
$$

If $AP$ is of length $s$, then $PB$ is of length $sx$.
The probability that the cutting point is on $AP$ is $\frac{s}{s + sx} = \frac{1}{1 + x}$.
Since the probability of the cutting point being at the same distance from the other end $B$ is the same, the probability of the desired event is $\frac{2}{x + 1}$.
Therefore, the answer is $(E)$.