A coin is tossed three times, where

Solution:

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, $\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$