Q.A die is thrown once. Let $A = \{1, 3, 5\}$, $B = \{2, 4, 6\}$, $C = \{1, 2\}$. Determine, giving reasons, which of the pairs $(A,B)$, $(A,C)$, $(B,C)$ are mutually exclusive.

[!TLDR] Intersect each pair and check for the empty set. [!ANSWER] $(A,B)$ mutually exclusive; $(A,C)$ and $(B,C)$ are NOT.

$A=\{1,3,5\}$, $B=\{2,4,6\}$, $C=\{1,2\}$. $A \cap B$: no odd number is even, so $A \cap B = \phi$ — mutually exclusive. $A \cap C$: the only common element of $\{1,3,5\}$ and $\{1,2\}$ is $1$, so $A \cap C = \{1\} \ne \phi$ — not mutually exclusive. $B \cap C$: the only common element of $\{2,4,6\}$ and $\{1,2\}$ is $2$, so $B \cap C = \{2\} \ne \phi$ — not mutually exclusive. [!ANSWER] $A$ and $B$ are mutually exclusive; $A,C$ and $B,C$ are each NOT mutually exclusive.

Compute the intersection of every pair of event-sets explicitly and check whether it is empty.

A common mistake is checking only one pair (e.g. $A,B$) and assuming the same conclusion carries over to the other pairs — each pair must be checked separately, since $C$ shares an element with both $A$ and $B$.