$14 \cdot 16$ If 2.5 is randomly split into the sum of two non-negative numbers, such as $2.5=2.143+0.357$ or $2.5=\sqrt{3}+(2.5-\sqrt{3})$, and each number is then changed to the nearest integer, such as the first example's $2.143,0.357$ being changed to 2,0; the second example's $\sqrt{3},(2.5-\sqrt{3})$ being changed to 2,1. Then the probability that the final sum of the two integers is 3 is
(A) $\frac{1}{4}$.
(B) $\frac{2}{5}$
(C) $\frac{1}{2}$.
(D) $\frac{3}{5}$.
(E) $\frac{3}{4}$.
(38th American High School Mathematics Examination, 1987)
[Solution 1] For 2.5, according to the problem, decompose (into two non-negative numbers) and round (write these two numbers as the closest integers) the two numbers can only be the following scenarios:
$$
0,2 ; 1,1 ; 2,0 ; 2,1 ; 1,2 \text {. }
$$

Among those with a sum of 3, there are 2 cases, so the probability is $\frac{2}{5}$.
Therefore, the answer is $(B)$.
[Solution 2] For the decomposition of $x+y=2.5$, it can be represented by the following diagram (each vertical line intersects the two number axes at points representing a decomposition), the circled numbers are the numbers when rounding in each interval segment (there are five segments).

Obviously, the number decompositions that fit the problem are only 2 segments. The probability is $\frac{2}{5}$. Therefore, the answer is $(B)$.