A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(ii) E: at least two heads, F: at most two heads
(iii) E: at most two tails, F: at least one tail
If a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space has 8 elements.
(i) E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
$\therefore \mathrm{E} \cap \mathrm{F}=\{\mathrm{HHH}\}$
$P(F)=\frac{2}{8}=\frac{1}{4}$ and $P(E \cap F)=\frac{1}{8}$
$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{4}{8}=\frac{1}{2}$
(ii) E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
$\therefore \mathrm{E} \cap \mathrm{F}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$
Clearly, $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{3}{8}$ and $\mathrm{P}(\mathrm{F})=\frac{7}{8}$
$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}$
(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}
F = {HHT, HTT, HTH, THH, THT, TTH, TTT}
$\therefore \mathrm{E} \cap \mathrm{F}=\{\mathrm{HHT}, \mathrm{HTT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{THT}, \mathrm{TTH}\}$
$P(F)=\frac{7}{8}$ and $P(E \cap F)=\frac{6}{8}$
Therefore, $\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$