Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black.

Let $C_1$ be the card with both sides red, $C_2$ be the card with both sides black, and $C_3$ be the card with one side red and one side black.

There are a total of 6 sides: $R_1, R_2$ (from $C_1$), $B_1, B_2$ (from $C_2$), and $R_3, B_3$ (from $C_3$).

The event $A$ is that the upper side of the chosen card is red. The possible outcomes for the upper side being red are $R_1, R_2, R_3$. Thus, $P(A) = \dfrac{3}{6} = \dfrac{1}{2}$.

The event $B$ is that the other side of the chosen card is black. We are looking for the conditional probability $P(B|A) = \dfrac{P(A \cap B)}{P(A)}$.

The event $A \cap B$ occurs if the upper side is red and the other side is black. This only happens if the card chosen is $C_3$ and the side $R_3$ is facing up. There is only 1 such outcome out of the 6 total sides.

Therefore, $P(A \cap B) = \dfrac{1}{6}$.

The conditional probability is $P(B|A) = \dfrac{1/6}{3/6} = \dfrac{1}{3}$.

Answer: $\dfrac{1}{3}$