Question
A coin is tossed thrice. If $$E$$ be the event of showing at least two heads and $$F$$ be the event of showing head in the first throw, then find $$P\left( {\frac{E}{F}} \right).$$
A.
$$\frac{4}{3}$$
B.
$$\frac{3}{4}$$
C.
$$\frac{1}{4}$$
D.
$$\frac{1}{2}$$
Answer :
$$\frac{3}{4}$$
Solution :
$$\eqalign{
& S = \left\{ {HHH,\,HHT,\,HTH,\,HTT,\,THH,\,THT,\,TTH,\,TTT} \right\} \cr
& E = \left\{ {HHH,\,HHT,\,HTH,\,THH} \right\} \cr
& F = \left\{ {HHH,\,HHT,\,HTH,\,HTT} \right\} \cr
& E \cap F = \left\{ {HHH,\,HHT,\,HTH} \right\} \cr
& n\left( {E \cap F} \right) = 3,\,n\left( F \right) = 4 \cr
& \therefore \,{\text{Required probability }} = P\left( {\frac{E}{F}} \right) = \frac{{n\left( {E \cap F} \right)}}{{n\left( F \right)}} = \frac{3}{4} \cr} $$