A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The prob

1.	A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is A. 0.39 B. 0.25 C. 0.11 D. none of these
Answer» Given, P(A) = 0.25 and P(B) = 0.5 Also P(A ∩ B) = 0.14 We have to find P(A’∩B’) By De Morgan’s theorem we know that: P(A’∩B’) = P(A∪B)’ We know that P(A∪B) = P(A) + P(B) – P(A∩B) ∴ P(A∪B) = 0.25 + 0.5 – 0.14 = 0.61 ∴ P(A’∩B’) = P(A∪B)’ = 1 - P(A∪B) = 1 – 0.61 = 0.39 As our answer matches with option (a) ∴ Option (a) is the only correct choice.

A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is A. 0.39 B. 0.25 C. 0.11 D. none of these

Answer»

Given, P(A) = 0.25 and P(B) = 0.5

Also P(A ∩ B) = 0.14

We have to find P(A’∩B’)

By De Morgan’s theorem we know that:

P(A’∩B’) = P(A∪B)’

We know that P(A∪B) = P(A) + P(B) – P(A∩B)

∴ P(A∪B) = 0.25 + 0.5 – 0.14 = 0.61

∴ P(A’∩B’) = P(A∪B)’ = 1 - P(A∪B)

= 1 – 0.61 = 0.39

As our answer matches with option (a)

∴ Option (a) is the only correct choice.

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A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is A. 0.39 B. 0.25 C. 0.11 D. none of these

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