A) dependent
B) independent
C) mutually exclusive
D) None of the above
Correct Answer: B
Solution :
We have \[P(A\cup B)=\frac{5}{6},(A\cap B)=\frac{1}{3},P(B)=\frac{1}{2}\] Now, \[P(B)=1-P(\bar{B})\] \[=1-\frac{1}{2}=\frac{1}{2}\] Also by addition theorem, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow \]\[P(A)=P(A\cup B)+P(A\cap B)-P(B)\] \[\Rightarrow \]\[P(A)=\frac{5}{6}+\frac{1}{3}-\frac{1}{2}\] \[\Rightarrow \]\[P(A)=\frac{2}{3}\] \[\therefore \] \[P(A)P(B)=\frac{2}{3}.\frac{1}{2}=\frac{1}{3}\] Thus, \[P(A\cap B)=P(A)P(B)\] Hence, A and B are independent events.You need to login to perform this action.
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