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Chapter 13 Class 12 Probability

Ex 13.1, 12 - If a family has two children. what conditional

Ex 13.1, 12 - Chapter 13 Class 12 Probability - Part 2
Ex 13.1, 12 - Chapter 13 Class 12 Probability - Part 3

Ex 13.1, 12 - Chapter 13 Class 12 Probability - Part 4 Ex 13.1, 12 - Chapter 13 Class 12 Probability - Part 5

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Ex 13.1, 12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girlA family has 2 children Let girls be denoted by ‘g’ & boys be denoted by ‘b’ S = {(g, g), (g, b), (b, g), (b, b)} We need to find the Probability that both the children are girls, given that the youngest is a girl. Let F : youngest child in a girl E : both the children are girls We need to find P(E|F) Also, E ∩ F = {(g, g)} So, P(E ∩ F ) = 1/4 Now, P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹)) = (1/4)/(1/2) E = {(g, g)} P(E) = 1/4 F = {(g, g), (b, g)} P(F) = 2/4=1/2 = 1/4 × 2/1 = 𝟏/𝟐 Ex 13.1, 12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (ii) at least one is a girl? S = {(g, g), (g, b), (b, g), (b, b)} We need to find the probability that the children are girls, given that at least one is a girl. F : at least one child is a girl E : both the children are girls E = {(g, g)} P(E) = 1/4 F = {(g, g), (g, b), (b, g)} P(F) = 3/4 Also, E ∩ F = {(g, g)} So, P(E ∩ F ) = 1/4 Now, P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹)) = (1/4)/(3/4) = 1/4 × 4/3 = 𝟏/𝟑

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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.