Three faces of a die are black, two faces are white and one face is...

Total number of faces on the die is $6$.

Number of black faces is $3$.
Probability of getting a black face in the first toss is $P(B) = \dfrac{3}{6} = \dfrac{1}{2}$.

Number of white faces is $2$.
Probability of getting a white face in the second toss is $P(W) = \dfrac{2}{6} = \dfrac{1}{3}$.

Number of red faces is $1$.
Probability of getting a red face in the third toss is $P(R) = \dfrac{1}{6}$.

Since the tosses are independent events, the required probability is the product of the individual probabilities.

Required probability $= P(B) \times P(W) \times P(R) = \dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{1}{6} = \dfrac{1}{36}$.

Answer: $1/36$