Q.A coin is tossed twice. Let $A =$ 'at least one head appears' and $B =$ 'no head appears'. Show that $A$ and $B$ are mutually exclusive and exhaustive.

[!TLDR] $A$ and $B$ are complementary events here, so both properties hold automatically. [!ANSWER] $A\cap B=\phi$ and $A\cup B={ …

$S = \{HH, HT, TH, TT\}$. $A$ = 'at least one head' $=\{HH,HT,TH\}$. $B$ = 'no head' $=\{TT\}$. Checking mutual exclusivity: $A$ contains no outcome with zero heads and $B$ contains only the zero-heads outcome, so $A \cap B = \phi$. Checking exhaustiveness: combining $A$ and $B$ gives $\{HH,HT,TH\} \cup \{TT\} = \{HH,HT,TH,TT\} = S$. [!ANSWER] $A \cap B = \ …

Recognise that 'at least one head' and 'no head' are complementary descriptions of the same sample space, then verify $A\cap …

A common mistake is missing that $B$ is precisely $A'$ (the complement of $A$) and re-deriving from scratch instead of using that every event and its …