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A,B,C are events such that `P_(r )(A)=0.3, P_(r )(B)=0.4,P_(r )(C )=0.8` `P_(r ) (AB)=0.08, P_(r )(AC )=0.28` and `P_(r )(B)=0.4, P_(r )(C )=0.8` If `P_(r )(A cup B cup C) ge 0.75`, then show that `P_(r )(BC )` lies in the interval [0.23,0.48]. Select the correct answer from above options

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Correct Answer - A We know that , P(A)+P(B)+P(C)-P(A∩B)-P(B∪C)-P(C∩A)+P(A∩B∩C)=P(A∪B∪C) ⇒0.3+0.4+0.8-{0.8+0.28+P(BC)}+0.09=P(A∪B∪C) ⇒1.23-P(BC)=P(A∪B∪C) where ,0.75≤P(A∪B∪B)≤1 ⇒0.75≤1.23-P(BC)≤1 ⇒-0.48≤-P(BC)≤-0.23 ⇒0.23≤P(BC)≤0.48

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