A) mutually exclusive and independent
B) independent but not equally likely
C) equally likely but not independent
D) equally likely and mutually exclusive
Correct Answer: B
Solution :
Given, \[P(\overline{A\cup B})=\frac{1}{6},P(\overline{A\cap B})=\frac{1}{4},P(\overline{A})=\frac{1}{4}\] \[\because \] \[P(\overline{A\cup B})=\frac{1}{6}\] \[\Rightarrow \] \[1-P(A\cup B)=\frac{1}{6}\] \[\Rightarrow \]\[1-P(A)-P(B)+P(A\cap B)=\frac{1}{6}\] \[\Rightarrow \] \[P(\overline{A})-P(B)+P(A\cap B)=\frac{1}{6}\] \[\Rightarrow \] \[-P(B)=\frac{1}{6}-\frac{1}{4}-\frac{1}{4}\] \[\Rightarrow \] \[P(B)=\frac{1}{3}\] and \[P(A)=\frac{3}{4}\] Clearly, \[P(A\cap B)=P(A).P(B)\] So, the events A and B are independent events but not equally likely.
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