Q.In the experiment of throwing a die once, let $A = \{1, 2, 3\}$ and $B = \{4, 5, 6\}$. Show that $A$ and $B$ are both mutually exclusive and exhaustive.

[!TLDR] Check $A\cap B=\phi$ (mutually exclusive) and $A\cup B=S$ (exhaustive). [!ANSWER] $A\cap B=\phi$ and $A\cup B=\{1,2,3,4,5,6\}=S$; both properties hold.

$S = \{1,2,3,4,5,6\}$, $A=\{1,2,3\}$, $B=\{4,5,6\}$. For mutual exclusivity, check $A \cap B$: the two sets share no common element, so $A \cap B = \phi$ — mutually exclusive. For exhaustiveness, check $A \cup B$: combining all elements of $A$ and $B$ gives $\{1,2,3,4,5,6\}$, which is exactly $S$ — so $A \cup B = S$, exhaustive. [!ANSWER] $A \cap B = \phi$ (mutually exclusive) and $A \cup B = S$ (exhaustive); $A$ and $B$ together partition $S$.

Test mutual exclusivity via $A\cap B$ and exhaustiveness via $A\cup B$ separately, as they are independent conditions.

A common mistake is proving only one of the two properties and assuming the other follows automatically — always verify both $A\cap B=\phi$ and $A\cup B=S$ explicitly.