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There are 2 independent event A and B such that, P(A ∪ B) = 0.72 and P(B') = 0.32. Find the probability P(A')
1. 0.225
2. 0.315
3. 0.875
4. 0.725

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Correct Answer - Option 3 : 0.875

Concept:

For any tow independent events A and B, if P(A) and P(B) are their probability of occurring then:

  • P(A ∩ B) = P(A) × P(B)
  • P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
  • P(A') = 1 - P(A)
  • P(B') = 1 - P(B)

 

Calculation:

Given P(B') = 0.32 and P(A∪B) = 0.72

Now P(B') = 1 - P(B)

⇒ P(B) = 1 - 0.32 = 0.68

Let P(A) = x

∵ P(A ∩ B) = P(A) × P(B)

⇒ P(A ∩ B) = 0.68 x

As P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ 0.72 = x + 0.68 - 0.68 x

⇒ 0.04 = 0.32 x

⇒ x = 0.125

∴ P(A) = 0.125

P(A') = 1 - P(A)

P(A') = 1 - 0.125

P(A') = 0.875

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