# Ex 13.1, 9 - Chapter 13 Class 12 Probability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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

Chapter 13 Class 12 Probability

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.