Given P(A \(\cup\) B)’ = 1/6, P(A \(\cap\) B) = 1/4 and P(A)’ = 1/4
\(\therefore\) P(A \(\cup\) B) = 1 – P(A \(\cup\) B)’
= 1 – \(1\over 6\) = \(5\over 6\)
and P(A) = 1 – P(A)’ = 1 – \(1\over 4\) = \(3\over 4\)
P(A \(\cup\) B) = P(A) + P(B) – P(A \(\cap\) B)
\(\implies\) \(5\over 6\) = \(3\over 4\) + P(B) – \(1\over 4\)
P(B) = \(1\over 3\)
\(\implies\) A and B are not equally likely,
Also, P(A \(\cap\) B) = P(A).P(B) = \(1\over 4\)
So, events are independent.
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