Q.Three fair coins are tossed simultaneously. Find the probability of getting exactly two heads.

List the sample space of three coins (or use the multiplication principle), then count the outcomes with exactly two heads.

Each of the 3 coins independently shows H or T, so by the multiplication principle (§1), $n(S) = 2 \times 2 \times 2 = 8$.

Listing the full sample space: $S=\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

'Exactly two heads' means precisely two of the three coins show H and the remaining one shows T. Scanning $S$: $HHT$ (coins 1,2 heads), $HTH$ (coins 1,3 heads), $THH$ (coins 2,3 heads) each satisfy this — no other outcome in $S$ has exactly two heads. So $E=\{HHT, HTH, THH\}$, $n(E)=3$.

By the classical definition (§5): $P(E) = \dfrac{n(E)}{n(S)} = \dfrac{3}{8}$. …

Alternative — via the multiplication principle for the count of favourable outcomes: choose which ONE of the 3 coins shows Tails (3 ways), with the other two automatically showing …

A common error is confusing 'exactly two heads' with 'at least two heads' — the latter would also include $HHH$ (three heads), giving $n(E)=4$ and $P=\tfrac48=\tfrac12$ instead of the correct $\tfrac38$; always re …