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Ex 13.1, 9 - Chapter 13 Class 12 Probability
Last updated at Aug. 16, 2023 by Teachoo
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
Serial order wise
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 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
