Ex 14.2

Chapter 14 Class 11 Probability
Serial order wise

### Transcript

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

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.