1. Chapter 13 Class 12 Probability
  2. Serial order wise
  3. Ex 13.2

Ex 13.2, 9 - Chapter 13 Class 12 Probability

Last updated at Dec. 8, 2016 by

Ex 13.2, 9 - If P(A) = 1/4, P(B)=1/4, find P(not A and not B) - Basic Probability

Slide21.JPG
Slide22.JPG

  1. Chapter 13 Class 12 Probability
  2. Serial order wise

    3. Ex 13.2

    Ex 13.3 →

Transcript

Ex 13.2, 9 (Method 1) If A and B are two events such that P (A) = 1﷮4﷯ , P (B) = 1﷮4﷯ and P(A ∩ B) = 1﷮8﷯ , find P (not A and not B). Given, P(A) = 1﷮4﷯ , P(B) = 1﷮2﷯ & P(A ∩ B) = 1﷮8﷯ P(not A & not B) = 1 – P(A or B) = 1 – P(A ∪ B) = 1 – [P(A) + P(B) – P(A ∩ B)] = 1 – 1﷮4﷯ + 1﷮2﷯ – 1﷮8﷯﷯ = 1 – 2﷮8﷯ + 4﷮8﷯ – 1﷮8﷯﷯ = 1 – 5﷮8﷯ = 𝟑﷮𝟖﷯ Ex 13.2, 9 (Method 2) If A and B are two events such that P (A) = 1﷮4﷯ , P (B) = 1﷮4﷯ and P(A ∩ B) = 1﷮8﷯ , find P (not A and not B). Given, P(A) = 1﷮4﷯ , P(B) = 1﷮2﷯ & P(A ∩ B) = 1﷮8﷯ P(not A & not B) = P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A∪B) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 1﷮4﷯ + 1﷮2﷯ – 1﷮8﷯ = 2﷮8﷯ + 4﷮8﷯ – 1﷮8﷯ = 5﷮8﷯ ∴ P(not A & not B) = 1 – 5﷮8﷯ = 𝟑﷮𝟖﷯

Chapter 13 Class 12 Probability
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.