P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∪C)
⇒α=1.4−p(A∩B)−β
⇒α+β=1.4−p(A∩B)....(1)
again
P(A∪B)=P(A)+P(B)−P(A∩B)
⇒P(A∩B)=⋅2......(2)
by (1) and (2)
α=1.2−β
now
0.85≤α≤0.95
⇒0.85≤1.2−β≤0.95⇒β∈[0.25,0.35]
The probabilities of three events A,B and C are given 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,P(B∩C)=β and P(A∪B∪C)=α, where 0.85≤α≤0.95, then β lies in the interval :
Held on 6 Sept 2020 · Verified 6 Jul 2026.
[0.35,0.36]
[0.25,0.35]
[0.20,0.25]
[0.36,0.40]
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