Question

A coin is tossed three times, where
(i) $E$ : head on third toss, $F$ : heads on first two tosses
(ii) $E$ : at least tow heads, $F$ : at most two heads
(iii) $E$ : at most two tails, $F$ : at least one tail
Determine $P\left(E|F\right)$

• A. $0.42,0.50,0.85$
• B. $0.50,0.42,0.85$
• C. $0.85,0.42,0.30$
• D. $.0.42,0.46,0.47$

Answer 1

Answer: (B)

$0.50,0.42,0.85$

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) $E=\{HHH,HTH,THH,TTH\}$
$F=\{HHH,HHT\}$
$\therefore E\cap F=\left\{HHH\right\}$
$P\left(F\right)=\frac{2}{8}=\frac{1}{4}$ and $P(E\cap F)=\frac{1}{8}$
$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{4}{8}=\frac{1}{2}=0.50$
(ii) $E=\{HHH,HHT,HTH,THH\}$
$F=\{HHT,HTH,HTT,THH,THT,TTH,TTT\}$
$\therefore E\cap F=\{HHT,HTH,THH\}$
Clearly, $P(E\cap F)=\frac{3}{8}$ and $P\left(F\right)=\frac{7}{8}$
$P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}=0.42$
(iii) $E=\{HHH,HHT,HTT,HTH,THH,THT,TTH\}$
$F=\{HHT,HTT,HTH,THH,THT,TTH,TTT\}$
$\therefore E\cap F=\{HHT,HTT,HTH,THH,THT,TTH\}$
$P\left(F\right)=\frac{7}{8}$ and $P(E\cap F)=\frac{6}{8}$
Therefore, $P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}=0.85$