In a throw of three dice, what is the probability of getting one...

The total number of outcomes when three dice are thrown is $6 \times 6 \times 6 = 216$.

The numbers on a die are $1, 2, 3, 4, 5, 6$.

Prime numbers are $2, 3, 5$ (total $3$).

Composite numbers are $4, 6$ (total $2$).

The number which is neither prime nor composite is $1$ (total $1$).

The number of ways to get one prime, one composite, and one neither prime nor composite is the number of permutations of these three types among the three dice multiplied by the number of choices for each type.

Number of favorable outcomes $= 3! \times (3 \times 2 \times 1) = 6 \times 6 = 36$.

Required probability $= \dfrac{36}{216} = \dfrac{1}{6}$.

Answer: $1/6$