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 :
\[P(\overline{A\cup B})\,=\frac{1}{6}\Rightarrow \,1-6(P\cup B)\,=\frac{1}{6}\] \[\Rightarrow \,\,(1-P(A))\,-P(B)\,+P(A\cap B)\,=\frac{1}{6}\] \[\Rightarrow \,P(\overline{A})\,-P(B)\,+P(A\cap B)\,=\frac{1}{6}\] \[\Rightarrow \,\frac{1}{4}\,-P(B)\,+\frac{1}{4}\,=\frac{1}{6}\] \[\Rightarrow P(B)=\frac{1}{3}\,and\,P(A)=1-\frac{1}{4}=\frac{3}{4}\] As \[P(A\cap B)\,=P(A)\,P(B)\] So, events A and B are independent events but they are not equally likely.You need to login to perform this action.
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