Events \(A\) and \(B\) are such that \(P(A)=\frac{1}{2},P(B)=\frac{7}{12}\) and (not \(A\) or not \(B\))\(=\frac{1}{4}\).State whether \(A\) and \(B\) are independent.
Solution and Explanation
\(\implies P(A'\cup B')=\frac{1}{4}\)
\(\implies P\bigg((A\cap B)'\bigg)=\frac{1}{4}\)
\(⇒\frac{1}{4}=1-P(A∩B)\)
\(⇒P(A∩B)=\frac{3}{4}\)
However, \(P(A).P(B)\)\(=\frac{1}{2}×\frac{7}{12}=\frac{7}{24}\)
\(\therefore\) \(P(A∩B)≠P(A).P(B)\) ,i.e., \(A\) and \(B\) are not independent.
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Concepts Used:
Independent Events
Independent Events are those events that are not dependent on the occurrence or happening of any other event. For instance, if we flip a dice and get 2 as the outcome, and if we flip it again and then get 6 as the outcome. In Both cases, the events have different results and are not dependent on each other.
All the events that are not dependent on the occurrence and nonoccurrence are denominated as independent events. If Event 1 does not depend on the occurrence of Event 2, then both Events 1 and 2 are independent Events.
Two Events: Event 1 and Event 2 are independent if,
P(2|1) = P (2) given P (1) ≠ 0
and
P (1|2) = P (1) given P (2) ≠ 0
Two events 1 and 2 are further independent if,
P(1 ∩ 2) = P(1) . P (2)