Q.Two fair coins are tossed together. Let $A =$ 'getting at least one head' and $B =$ 'getting no head'. Examine whether $A$ and $B$ are mutually exclusive and whether they are collectively exhaustive.

List the sample space of two coins, identify $A$ and $B$ as sets, then check their intersection and union.

$S=\{HH,HT,TH,TT\}$. $A=\{HH,HT,TH\}$, $B=\{TT\}$. $A \cap B=\emptyset$ and $A \cup B = S$, so both mutually exclusive and collectively exhaustive.

From §1/Q2, the sample space for two coins is $S=\{HH,HT,TH,TT\}$, $n(S)=4$.

$A=$ 'at least one head' means one or both coins show Heads: $A=\{HH,HT,TH\}$ (3 sample points).

$B=$ 'no head' means neither coin shows Heads (both Tails): $B=\{TT\}$ (1 sample point).

Mutual exclusivity: $A \cap B$ — is any outcome in BOTH 'at least one head' and 'no head'? No outcome can satisfy both simultaneously, so $A \cap B = \emptyset$ — mutually exclusive.

Collective exhaustiveness: $A \cup B = \{HH,HT,TH\} \cup \{TT\} = \{HH,HT,TH,TT\} = S$ — collectively exhaustive.

Why this makes sense: 'at least one head' and 'no head' are exact opposites (complementary events, §5) — for any pair of complementary events, they are always automatically both mutually exclusive and collectively exhaustive, since together they cover $S$ with no possible overlap.

Alternative — recognizing $B$ as the complement of $A$ directly: since $B = \text{not } A$ (no head is exactly the opposite of at least one head), any event and its complement are automatically mutually exclusive and collectively exhaustive, without needing to list and compare sets explicitly.

A common error is mis-listing $A$ as only $\{HH\}$ (misreading 'at least one head' as 'exactly two heads' or 'all heads') instead of correctly including $HT$ and $TH$ as well — 'at least one' must include every outcome with one OR more heads, not just the maximum case.