Ex 14.2, 13 - Chapter 14 Class 11 Probability

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Ex 14.2, 13 Fill in the blanks in following table: We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values P(A ∪ B) = 1﷮3﷯ + 1﷮5﷯ – 1﷮15﷯ P(A ∪ B) = 5 + 3 − 1﷮15﷯ P(A ∪ B) = 7﷮15﷯ Hence P(A ∪ B) = 𝟕﷮𝟏𝟓﷯ Ex 14.2, 13 Fill in the blanks in following table: We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 0.6 = 0.35 + P(B) – 0.25 0.6 = 0.35 – 0.25 + P(B) 0.6 = 0.10 + P(B) 0.6 – 0.10 = P(B) 0.5 = P(B) P(B) = 0.5 Hence P(B) = 0.5 Ex 14.2, 13 Fill in the blanks in following table: We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 0.7 = 0.5 + 0.35 – P(A ∩ B) 0.7 = 0.85 – P(A ∩ B) P(A ∩ B) = 0.85 – 0.7 P(A ∩ B) = 0.15 Hence, P(A ∩ B) = 0.15

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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