Ex 14.2, 12 - Check whether P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 are - Ex 14.2

part 2 - Ex 14.2, 12 (ii) - Ex 14.2 - Serial order wise - Chapter 14 Class 11 Probability

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Ex 14.2, 12 Check whether the following probabilities P(A) and P(B) are consistently defined P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 P(A) & P(B) are consistently defined if P(A ∩ B) < P(A) & P(A ∩ B) < P(B) P(A ∪ B) > P(A) & P(A ∪ B) > P(B) Given P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 Here, P(A ∩ B) > P(A). Hence, P(A) and P(B) are not consistently defined. Ex 14.2, 12 Check whether the following probabilities P(A) and P(B) are consistently defined (ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 P(A) & P(B) are consistently defined if P(A ∩ B) < P(A) & P(A ∩ B) < P(B) P(A ∪ B) > P(A) & P(A ∪ B) > P(B) Given P(A) = 0.5, P(B) = 0.4 P(A ∪ B) = 0.8 Here, P(A ∪ B) > P(A) & P(A ∪ B) > P(B) Finding P(A ∩ B) We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 0.8 = 0.5 + 0.4 – P(A ∩ B) 0.8 = 0.9 – P(A ∩ B) P(A ∩ B) = 0.9 – 0.8 P(A ∩ B) = 0.1 So, P(A ∩ B) < P(A) & P(A ∩ B) < P(B) Since both conditions are satisfied, Hence, P(A) and P(B) are consistently defined.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo