Let A, B and C be three events such that `P(A)=0.3, P(B)=0.4, P(C )=0.8, P(A cap B)=0.08, P(A cap C) =0.28, P(A cap B cap C)=0.09`. If `P(A cup B cup

Question

Let A, B and C be three events such that P(A)=0.3,P(B)=0.4,P(C)=0.8,P(A∩B)=0.08,P(A∩C)=0.28,P(A∩B∩C)=0.09 P ( A ) = 0.3 , P ( B ) = 0.4 , P ( C ) = 0.8 , P ( A ∩ B ) = 0.08 , P ( A ∩ C ) = 0.28 , P ( A ∩ B ∩ C ) = 0.09 . If P(A∪B∪C)≥0.75 , then show that P(B∩C) satisfies A. P(B∩C)≤0.23 B. P(B∩C)≤0.48 C. 0.23≤P(B∩C)≤0.48 D. 0.23≤P(B∩C)≤0.48 Select the correct answer from above options

Answer 1

Correct Answer - C We have that the probability of occurrence of an event is always less than or equal to 1 and it is given that P(A∪B∪C)≥0.75 . ∴0.75≤P(A∪B∪C)≤1 ⇒0.75≤P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(A∩C)+P(A∩B∩C)≤1 ⇒ 0.75 ≤ P ( A ) + P ( B ) + P ( C ) − P ( A ∩ B ) − P ( B ∩ C ) − P ( A ∩ C ) + P ( A ∩ B ∩ C ) ≤ 1 ⇒0.75≤0.3+0.4+0.8-0.08-P(B∩C)-0.28+0.09≤1 ⇒0.75≤1.59-0.36-P(B∩C)≤1 ⇒0.75≤1.23-P(B∩C)≤1 ⇒-0.48≤-P(B∩C)≤-0.23⇒0.23≤P(B∩C)≤0.48