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 :

Question

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 :, 0 . 35 , 0 . 36, 0 . 25 , 0 . 35, 0 . 20 , 0 . 25, 0 . 36 , 0 . 40

MARKS App · Accepted Answer

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

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