A) mutually exclusive and independent.
B) equally likely but not independent.
C) independent but not equally likely.
D) independent and equally likely.
Correct Answer: C
Solution :
\[P(\left( \overline{A\cup B} \right)=\frac{1}{6}\] \[1-P(A\cup B)=\frac{1}{6}\] \[P(A\cup B)=\frac{5}{6}\] \[P(A\cap B)=\frac{1}{4}\] \[P(\overline{A})=\frac{1}{4}\] \[P(A)=\frac{3}{4}\] \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\frac{5}{6}=\frac{3}{4}+P(B)-\frac{1}{4}\] \[P(B)=\frac{5}{6}-\frac{1}{2}\] \[P(B)=\frac{1}{3}\] A of B are Independent because \[P(A\cap B)=P(A)P(B).A\]and B has different probability so if is not equally likely.You need to login to perform this action.
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