The probabilities of three events A , B and C are given P A = 0 . 6 ,   P B = 0 . 4 and P C = 0 . 5 .…

The probabilities of three events A,B and C are given PA=0.6, PB=0.4 and PC=0.5. If PA∪B=0.8, PA∩C=0.3, PA∩B∩C=0.2, PB∩C=β and PA∪B∪C=α, where 0.85≤α≤0.95, then β lies in the interval :
  1. 0.35, 0.36
  2. 0.25, 0.35
  3. 0.20, 0.25
  4. 0.36, 0.40

Solution

PA∪B∪C=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(C∩A)+PA∩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

Asked in: JEE Main 2020 (06 Sep Shift 2)

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