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:
\[\mathbb{P}(A) = 1 − \mathbb{P}(A^c) =\frac{11}{36}\]