A) Independent but not equally likely
B) Mutually exclusive and independent
C) Equally likely and mutually exclusive
D) Equally likely but not independent
Correct Answer: A
Solution :
\[P\left( \overline{A\cup B} \right)=\frac{1}{6};\,\,P\left( A\cap B \right)=\frac{1}{4}\], \[P\left( {\bar{A}} \right)=\frac{1}{4}\Rightarrow P\left( A \right)=\frac{3}{4}\], \[P\left( \overline{A\cup B} \right)=1-P\left( A\cup B \right)=1-P\left( A \right)-P\left( B \right)+P\left( A\cap B \right)\] Þ \[\frac{1}{6}=\frac{1}{4}-P\left( B \right)+\frac{1}{4}\]Þ\[P\left( B \right)=\frac{1}{3}\]. Since \[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\] and \[P\left( A \right)\ne P\left( B \right)\] \ A and B are independent but not equally likely.You need to login to perform this action.
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