Ex 14.2, 17 - P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16 - Ex 14.2

part 2 - Ex 14.2, 17 - Ex 14.2 - Serial order wise - Chapter 14 Class 11 Probability
part 3 - Ex 14.2, 17 - Ex 14.2 - Serial order wise - Chapter 14 Class 11 Probability

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Ex 14.2, 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 14.2, 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 14.2, 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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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