Question
If $${E_1}$$ and $${E_2}$$ are two events such that $$P\left( {{E_1}} \right) = \frac{1}{4},\,P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{1}{2}$$ and $$P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = \frac{1}{4},$$ then choose the incorrect statement.
A.
$${E_1}$$ and $${E_2}$$ are independent
B.
$${E_1}$$ and $${E_2}$$ are exhaustive
C.
$${E_2}$$ is twice as likely to occur as $${E_1}$$
D.
Probabilities of the events $${E_1} \cap {E_2},\,{E_1}$$ and $${E_2}$$ are in G.P.
Answer :
$${E_1}$$ and $${E_2}$$ are exhaustive
Solution :
$$\eqalign{
& P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}} \cr
& \Rightarrow \frac{1}{2} = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{\frac{1}{4}}} \cr
& \Rightarrow P\left( {{E_1} \cap {E_2}} \right) = \frac{1}{8} \cr
& = P\left( {{E_2}} \right).P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = P\left( {{E_2}} \right).\frac{1}{4} \cr
& \Rightarrow P\left( {{E_2}} \right) = \frac{1}{2} \cr
& {\text{Since, }}P\left( {{E_1} \cap {E_2}} \right) = \frac{1}{8} = P\left( {{E_1}} \right).P\left( {{E_2}} \right) \cr
& \Rightarrow {\text{events are independent}} \cr
& {\text{Also, }}P\left( {{E_1} \cup {E_2}} \right) = \frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8} \cr
& \Rightarrow {E_1}\,\,\& \,\,{E_2}\,{\text{are non exhaustive}}{\text{.}} \cr} $$