16. (6 points) If $2^{a} \times 3^{b} \times 5^{c} \times 7^{d}=252000$, then the probability that a three-digit number formed by randomly selecting 3 out of the natural numbers $a$, $b$, $c$, $d$ is divisible by 3 and less than 250 is $\qquad$ .
【Solution】Solution: First, decompose 252000 into prime factors as $7 \times 3^{2} \times 2^{5} \times 5^{3}$
$$
a=5, b=2, c=3, d=1 \text {. }
$$

There are a total of $\mathrm{A}_{4}^{3}=4 \times 3 \times 2=24$ three-digit numbers.
Numbers less than 250 with 1 as the leading digit are $123,125,132,135,152,153$, a total of 6. Among them, the numbers divisible by 3 are $123,132,153,135$.

Numbers starting with 2 are $213,215,231,235$, a total of 4. Among them, the multiples of 3 are 213,231, a total of 2.
The probability is $\frac{4+2}{24}=\frac{1}{4}$
Therefore, the answer is: $\frac{1}{4}$.