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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 middleMother, Father and son line up at random for a family picture So, Sample Space will be 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 that the father is in the middle. So, E: Son on one end F: Father in middle We need to find P(E|F) Event E E = {(M,F,S), (F,M,S),(S,M,F), (S,F,M)} P(E) = 4/6=2/3 Event F F = {(M, F, S), (S, F, M)} P(F) = 2/6=1/3 Also, E ∩ F = {(M,F,S), (S,F,M)} ∴ 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.