Q.Three coins are tossed together. Find the probability of getting

Three coins give $2^3=8$ equally likely outcomes; count those matching each event.

(i) $\dfrac{3}{8}$; (ii) $1-\dfrac{1}{8}=\dfrac{7}{8}$.

With three coins, $n(S)=2^3=8$ and

1. Exactly two heads. Favourable: $HHT, HTH, THH$ — $3$ outcomes, so

   $P=\frac{3}{8}.$

2. At least one head. The only outcome with no head is $TTT$, so by the complement rule

   $P(\text{at least one head})=1-P(TTT)=1-\frac{1}{8}=\frac{7}{8}.$

   Independent check for (i): the number of ways to place exactly $2$ heads among $3$ tosses is $\binom{3}{2}=3$, giving $\dfrac{3}{8}$ — matches the list.
   ✓Final answer

   (i) $P=\dfrac{3}{8}$; (ii) $P=\dfrac{7}{8}$.

Use $\binom{3}{r}/8$ for "exactly $r$ heads"; "at least one head" is best via the complement of the all-tails outcome.

For "at least one head", trying to add many cases and slipping up, instead of the one-step complement $1-\tfrac18$. For "exactly two heads", accidentally including $HHH$ (that is three heads, not two).