Question
For two mutually exclusive events $$A$$ and $$B,\,P\left( A \right) = 0.2$$ and $$P\left( {\overline A \cap B} \right) = 0.3.$$ What is $$P\left( {A\left| {\left( {A \cup B} \right)} \right.} \right)$$ equal to ?
A.
$$\frac{1}{2}$$
B.
$$\frac{2}{5}$$
C.
$$\frac{2}{7}$$
D.
$$\frac{2}{3}$$
Answer :
$$\frac{2}{5}$$
Solution :
$$\eqalign{
& {\text{Events }}A{\text{ and }}B{\text{ are mutually exclusive}}{\text{.}} \cr
& {\text{Hence,}}\,P\left( {A \cap B} \right) = \phi = 0 \cr
& \therefore \,P\left( {A \cup B} \right) + P\left( A \right) + P\left( B \right)......\left( 1 \right) \cr
& P\left( A \right) = 0.2\,\,\,\,\,\,\,\,\left[ {{\text{given}}} \right] \cr
& P\left( B \right) = P\left( {\overline A \cap B} \right) + P\left( {A \cap B} \right) \cr
& \Rightarrow P\left( B \right) = P\left( {\overline A \cap B} \right)\,\,\,\,\,\,\,\,\,\,\left[ {\because P\left( {A \cap B} \right) = 0} \right] \cr
& \Rightarrow P\left( B \right) = 0.3 \cr
& P\left( {A \cup B} \right) = 0.2 + 0.3 = 0.5 \cr
& P\left( {A\left| {\left( {A \cup B} \right)} \right.} \right) = \frac{{P\left( A \right)}}{{P\left( {A \cup B} \right)}} = \frac{{0.2}}{{0.5}} \cr
& \Rightarrow P\left( {A\left| {\left( {A \cup B} \right)} \right.} \right) = \frac{2}{5} \cr} $$