One way to calculate the probability of the event 'at least one number is a $6$' is to make use of the complement rule:

$$\mathbb{P}(A) = 1 − \mathbb{P}(A^c)$$

Here, $A^c$ is the event of getting no sixes in two rolls.

To calculate the value of $\mathbb{P}(A^c)$, we define events $B_1$ and $B_2$ as follows:

• $B_1=$ 'the first roll is not a $6$'
  
  
• $B_2=$ 'the second roll is not a $6$'
  
  

The probabilities of these events are:

• $\mathbb{P}(B_1) = \cfrac{5}{6}$
  
  
• $\mathbb{P}(B_2) = \cfrac{5}{6}$
  
  

The event of neither roll being a $6$ is equivalent to the combined event of getting no $6$ on the first roll AND getting no $6$ on the second roll:

$$A^c = B_1 \cap B_2$$
Since dice rolls are independent events, we can use the following formula to calculate the probability of the intersection of $B_1$ and $B_2$:
$$\mathbb{P}(A^c)=\mathbb{P}(B_1 \cap B_2) = \mathbb{P}(B_1) \cdot \mathbb{P}(B_2) = \frac{5}{6}\cdot \frac{5}{6}=\frac{25}{36}$$
Now that $\mathbb{P}(A^c)$ is known, $\mathbb{P}(A)$ can be calculated:
$$\mathbb{P}(A) = 1 − \mathbb{P}(A^c) =\frac{11}{36}$$