Question
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$
Answer :
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
Solution :
$$\eqalign{
& P\left( {\frac{{\overline A }}{{\overline B }}} \right) = \frac{{P\left( {\overline A \cap \overline B } \right)}}{{P\left( {\overline B } \right)}} \cr
& = \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\overline B } \right)}} \cr
& = \frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}} \cr} $$