# Misc 7

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B) Given P(A) = 0.54 , P(B) = 0.69 , P(A ∩ B) = 0.35 We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B) Putting values = 0.54 + 0.69 − 0.35 = 0.88 Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (ii) P(A′ ∩ B′) P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12 Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (iii) P(A ∩ B′) P(A ∩ B′) = P(A) − P(A ∩ B) Putting values = 0.54 − 0.35 = 0.19 Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (iv) P(B ∩ A′) P(B ∩ A′) = P(B) – P(A ∩ B) Putting values = 0.69 – 0.35 = 0.34

Chapter 16 Class 11 Probability

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.