Question
Let $$A$$ and $$B$$ be two events such that $$P\left( {\overline {A \cup B} } \right) = \frac{1}{6},P\left( {\overline {A \cap B} } \right) = \frac{1}{4}$$ and $$P\left( {\overline A } \right) = \frac{1}{4},$$ where $$\overline A $$ stands for the complement of the event $$A.$$ Then the events $$A$$ and $$B$$ are
A.
independent but not equally likely.
B.
independent and equally likely.
C.
mutually exclusive and independent.
D.
equally likely but not independent.
Answer :
independent but not equally likely.
Solution :
Given
$$\eqalign{
& P\left( {\overline {A \cup B} } \right) = \frac{1}{6} \cr
& \Rightarrow \,\,P\left( {A \cup B} \right) = 1 - \frac{1}{6} = \frac{5}{6} \cr} $$
$$\eqalign{
& P\left( {\overline A } \right) = \frac{1}{4} \cr
& \Rightarrow \,\,P\left( A \right) = 1 - \frac{1}{4} = \frac{3}{4} \cr} $$
We know
$$\eqalign{
& P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr
& \Rightarrow \,\,\frac{5}{6} = \frac{3}{4} + P\left( B \right) - \frac{1}{4}\,\,\left( {\because \,P\left( {A \cap B} \right) = \frac{1}{4}} \right) \cr
& \Rightarrow \,\,P\left( B \right) = \frac{1}{3} \cr} $$
$$\because \,\,P\left( A \right) \ne P\left( B \right)$$ so they are not equally likely.
Also $$P\left( A \right) \times P\left( B \right) = \frac{3}{4} \times \frac{1}{3}$$
$$ = \frac{1}{4} = P\left( {A \cap B} \right)$$
So $$A$$ & $$B$$ are independent.