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Question

# The probabilites of three events A,B and C are P(A)=0.6,P(B)=0.4 and P(C)=0.5. If P(A∪B)=0.8,P(A∩C)=0.3,P(A∩B∩C)=0.2 and P(A∪B∪C)≥0.85, then

A
0.2P(BC)0.35
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B
0.5P(BC)0.85
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C
0.1P(BC)0.35
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D
None of these
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Solution

## The correct option is C 0.2≤P(B∩C)≤0.35Given P(A)=0.6,P(B)=0.4 and P(C)=0.5. If P(A∪B)=0.8,P(A∩C)=0.3,P(A∩B∩C)=0.2 and P(A∪B∪C)≥0.85P(A∩B)=P(A)+P(B)−P(A∪B)=0.6+0.4−0.8=0.2P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩C)+P(A∩B∩C)−P(A∩B)−P(B∩C)P(A∪B∪C)=0.6+0.4+0.5−0.3+0.2−0.2−P(B∩C)⇒P(B∩C)=1.2−P(A∪B∪C) ...(1)∵0.85≤P(A∪B∪C)≤1⇒P(B∩C)≤1.2−0.85 [From (1)]and P(B∩C)≥1.2−1⇒0.2≤P(B∩C)≤0.35.

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