$\mathbb{P}(A \backslash B) = \cfrac{1}{6}$

The probabilities of events $A$ and $B$ are:

• $\mathbb{P}(A)=\cfrac{\text{'number of outcomes}\leq 3\text{'}}{\text{'total number of outcomes'}}=\cfrac{3}{6}$
• $\mathbb{P}(B) = \cfrac{\text{'number of odd outcomes'}}{\text{'total number of outcomes'}}=\cfrac{3}{6}$

In order to calculate the probability of the difference, first calculate the probability of the intersection $\mathbb{P}(A \cap B)$.

To determine how the intersection should be calculated, investigate whether events $A$ and $B$ are independent or not. If it is known that the outcome of the roll is a number $\leq 3$, then the probability of the roll being odd is $2$ out of $3$; namely $1$ and $3$, but not $2$:

$$\mathbb{P}(B|A) =\cfrac{2}{3}$$

This demonstrates that $\mathbb{P}(B) \neq \mathbb{P}(B|A)$, so $A$ and $B$ are not independent. Therefore, the multiplication rule should be applied:

$$\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B|A) =\cfrac{3}{6}\cdot \cfrac{2}{3}=\cfrac{2}{6}$$

Now, the probability of the difference of $A$ and $B$ can be calculated:

$$\mathbb{P}(A \backslash B) = \mathbb{P}(A) - \mathbb{P}(A \cap B) = \cfrac{3}{6}-\cfrac{2}{6}=\cfrac{1}{6}$$