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Ex 13.2, 7 - Chapter 13 Class 12 Probability (Term 2)
Last updated at May 18, 2022 by Teachoo
Ex 13.2
Ex 13.2, 1
Ex 13.2, 2
Ex 13.2, 3 Important
Ex 13.2, 4
Ex 13.2, 5
Ex 13.2, 6
Ex 13.2, 7 Important You are here
Ex 13.2, 8
Ex 13.2, 9 Important
Ex 13.2, 10 Important
Ex 13.2, 11 (i)
Ex 13.2, 11 (ii) Important
Ex 13.2, 11 (iii)
Ex 13.2, 11 (iv) Important
Ex 13.2, 12
Ex 13.2, 13 Important
Ex 13.2, 14 Important
Ex 13.2, 15 (i)
Ex 13.2, 15 (ii)
Ex 13.2, 15 (iii) Important
Ex 13.2, 16 Important
Ex 13.2, 17 (MCQ)
Ex 13.2, 18 (MCQ) Important
Chapter 13 Class 12 Probability (Term 2)
Serial order wise
Transcript
Ex 13.2, 7 (i) Given that the events A and B are such that P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Find p if they are (i) mutually exclusiveGiven, P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Given sets A & B are mutually exclusive, So, they have nothing in common ∴ P(A ∩ B) = 0 We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 3/5 = 1/2 + p – 0 3/5 – 1/2 = p (6 − 5)/10 = p 1/10 = p p = 𝟏/𝟏𝟎 Ex 13.2, 7 (ii) Given that the events A and B are such that P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Find p if they are (ii) independent.Since events A & B are independent, So, P(A ∩ B) = P(A) P(B) = 1/2 × p = 𝑝/2 Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 3/5 = 1/2 + p – 𝑝/2 3/2 – 1/2 = p – 𝑝/2 (6 − 5)/10 = 𝑝/2 1/10 = 𝑝/2 p = 2/10 p = 𝟏/𝟓
