Ex 16.3, 17 - P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16 - Using formulae of sets

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  1. Chapter 16 Class 11 Probability
  2. Serial order wise
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Ex 16.3, 17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), P(A) = 0.42 P(not A) = 1 – P(A) = 1 – 0.42 = 0.58 Ex 16.3, 17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (ii) P (not B) P(B) = 0.48 P(not B) = 1 – P(B) = 1 – 0.48 = 0.52 Ex 16.3, 17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (iii) P(A or B). P(A or B) = P(A ∪ B) We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values P(A ∪ B) = 0.42 + 0.48 – 0.16 = 0.74 Hence, P(A or B) = 0.74

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