Q.Two fair coins are tossed simultaneously. Write the sample space and find $n(S)$. Also write the event 'getting exactly one head' as a set, and state whether it is a simple or compound event.

Each of the two coins can independently land Head or Tail, so listing every combination gives the sample space; then pick out the outcomes matching 'exactly one head.'

$S = \{HH, HT, TH, TT\}$, $n(S) = 4$. 'Exactly one head' $= \{HT, TH\}$, which has 2 sample points, so it is a compound event.

Each coin independently lands H or T, so by the multiplication principle (§1) the combined sample space has $n(S) = 2 \times 2 = 4$ sample points.

Listing every pair (first coin's result, second coin's result): $S = \{HH, HT, TH, TT\}$.

The event 'exactly one head' means precisely one of the two coins shows Heads and the other shows Tails — scanning $S$, this matches $HT$ (first coin Head, second Tail) and $TH$ (first coin Tail, second Head), giving the event $\{HT, TH\}$.

This event contains $2$ sample points, which is more than one — by the definition in §2, an event with two or more sample points is a compound event, formed here by combining the two simple events $\{HT\}$ and $\{TH\}$.

Alternative — using the multiplication principle directly to confirm $n(S)$: since each of the 2 coins has 2 outcomes, $n(S) = 2 \times 2 = 4$, matching the explicit listing without needing to write the pairs out first.

A frequent error is treating $HT$ and $TH$ as the 'same' outcome (both look like 'one head, one tail') and undercounting the sample space to only 3 outcomes $\{HH, \text{one-head}, TT\}$ — but the two coins are distinguishable by which one shows which face (or equivalently, by toss order), so $HT$ and $TH$ are genuinely different sample points.