A) \[P\,\left( \frac{A}{B} \right)=\frac{1}{2}\]
B) \[P\,\left( \frac{A}{A\cup B} \right)=\frac{5}{6}\]
C) \[P\,\left( \frac{A\cap B}{{A}'\cup {B}'} \right)=0\]
D) All of the above
Correct Answer: D
Solution :
\[P(A/B)=P(A)\] as independent event \[=\frac{1}{2}.\] \[P\{A/(A\cup B)\}=\frac{P[A\cap (A\cup B)]}{P(A\cup B)}\] {Since \[A\cap (A\cup B)=A\cap [A-B-A\cap B]\] \[=A-A\cap B-A\cap B=a\}\] \[\Rightarrow P\left( \frac{A}{A\cup B} \right)=\frac{P(A)}{P(A\cup B)}=\frac{\frac{1}{2}}{\frac{1}{2}-\frac{1}{5}-\frac{1}{10}}=\frac{\frac{1}{2}}{\frac{6}{10}}=\frac{5}{6}\] and similarly \[P\left( \frac{A\cap B}{{A}'\cup {B}'} \right)\].
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