When three coins are tossed once, the sample space $S$ is given by
$S=\{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\}$
Number of exhaustive cases $=8$
(i) There is only one favourable case $=\mathrm{HHH}$
$\therefore P(3 \text { heads })=\frac{1}{8}$
(ii) There are three favourable cases when two heads occur viz. HHT, HTH, THH.
$\therefore P($ exactly 2 heads $)=\frac{3}{8}$
(iii) At least 2 head $\Rightarrow 2$ or 3 heads
There are 4 favourable case $\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HHH}$
$\therefore \quad P($ at least 2 heads $)=\frac{4}{8}=\frac{1}{2}$
(iv) $P($ at most 2 heads $)=P($ not 3 heads $)$
$=1-P(3 \text { heads })=1-\frac{1}{8}=\frac{7}{8}$
(v) No head means all tails are obtained. There is only one favourable case $=T T T$.
$\therefore \quad P(\text { no head })=\frac{1}{8}$
(vi) There is only one favourable case TTT.
$\therefore P(3 \text { tails })=\frac{1}{8}$
(vii) There are 3 favourable cases $H T T, T H T, T T H$.
$\therefore P($ exactly 2 tails $)=\frac{3}{8}$
(viii) No Tail means all heads are obtained. There is only one favourable case $H H H$.
$\therefore P($ all heads $)=\frac{1}{8}$
(ix) At most two tails $\Rightarrow$ All the 3 tails do not occur 3 tails occur is 1 way.
$\therefore$ Probability that 3 tails appear $=\frac{1}{8}$ Probability that 3 tails do not appear $=1-\frac{1}{8}=\frac{7}{8}$
$\therefore$ Probability of getting at most two tails $=\frac{7}{8}$