Question

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.

Answer 1

If a coin is tossed 3 times, then the sample space S is

$S=HHH,HHT,HTH,HTT,THH,THT,TTH,TTT,n\left(S\right)=8$

(i) $E=HHH,HTH,THH,TTH$ and $F=HHH,HHT$

$\therefore E\cap F=HHH$

$P\left(F\right)=\frac{2}{8}=\frac{1}{8},P(E\cap F)=\frac{1}{8}$

Hence the probability in first case will be,

$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{1}{8}\times \frac{4}{1}=\frac{1}{2}$

(ii) $E=HHH,HHT,HTH,THH$

$F=HHT,HTH,HTT,THH,THT,TTH,TTT$

$\therefore E\cap F=HHT,HTH,THH$

Clearly, $P(E\cap F)=\frac{3}{8},P\left(F\right)=\frac{7}{8}$

Hence the required probability in the second case will be,

$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{3}{8}\times \frac{8}{7}=\frac{3}{7}$

(iii) $E=HHH,HHT,HTT,HTH,THH,THT,TTH$

$F=HHT,HTT,HTH,THH,THT,TTH,TTT$

$\therefore E\cap F=HHT,HTT,HTH,THH,THT,TTH$

$P\left(F\right)=\frac{7}{8}=\frac{1}{8},P(E\cap F)=\frac{6}{8}$

Hence the probability in third case will be,

$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{6}{8}\times \frac{8}{7}=\frac{6}{7}$