Ex 13.1, 9 - Chapter 13 Class 12 Probability (Term 2)
Last updated at May 25, 2021 by
Transcript
Ex 13.1, 9 Mother, father and son line up at random for a family picture E : son on one end, F : father in middle Mother, Father and son line up at random for a family picture Let mother be denoted by ‘M’ father be denoted by ‘F’ & son be denoted by ‘S’ So, S = { (M,F,S), (M,S,F), (F,M,S), (F,S,M), (S,M,F), (S,F,M) } We need to find the probability that the son is on one end, given the father is in the middle. F : father in middle E : son on one end We need to find P(E|F) Also, E ∩ F = {(M,F,S), (S,F,M)} So, P(E ∩ F ) = 26= 13 Now, P(E|F) = 𝑃(𝐸 ∩ 𝐹)𝑃(𝐹) = 13 13 = 1
Ex 13.1
Ex 13.1, 1
Ex 13.1, 2
Ex 13.1, 3 Important
Ex 13.1, 4
Ex 13.1, 5
Ex 13.1, 6 (i)
Ex 13.1, 6 (ii) Important
Ex 13.1, 6 (iii)
Ex 13.1, 7 (i)
Ex 13.1, 7 (ii)
Ex 13.1, 8
Ex 13.1, 9 You are here
Ex 13.1, 10 (a) Important
Ex 13.1, 10 (b) Important
Ex 13.1, 11
Ex 13.1, 12 Important
Ex 13.1, 13 Important
Ex 13.1, 14
Ex 13.1, 15
Ex 13.1, 16 (MCQ) Important
Ex 13.1, 17 (MCQ) Important
Chapter 13 Class 12 Probability (Term 2)
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.
