(B) The sample space $S$ consists of $36$ outcomes.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Since $B$ is the event of getting an odd number on the first die,its complement $B^{\prime}$ is the event of getting an even number on the first die.
Thus,$B^{\prime} = A$.
Now,$A \cap B^{\prime} = A \cap A = A$.
Since $A$ is not an empty set $(A \neq \phi)$,$A \cap B^{\prime} \neq \phi$.
Therefore,$A$ and $B^{\prime}$ are not mutually exclusive.
Thus,the given statement is false.