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Ex 13.1, 9 - Chapter 13 Class 12 Probability (Term 2)

Last updated at May 25, 2021 by

Ex 13.1, 9 - Mother, father, son line up at random for family - Ex 13.1

Ex 13.1, 9 - Chapter 13 Class 12 Probability - Part 2

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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 ) = 2﷮6﷯= 1﷮3﷯ Now, P(E|F) = 𝑃(𝐸 ∩ 𝐹)﷮𝑃(𝐹)﷯ = 1﷮3﷯﷮ 1﷮3﷯﷯ = 1

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