

Ex 13.5
Ex 13.5, 2 Deleted for CBSE Board 2022 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams
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Ex 13.5, 5 Deleted for CBSE Board 2022 Exams
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Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams
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Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams You are here
Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective? In each of the following, choose the correct answer:vIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i.e. Probability success = P then Probability failure = q = 1 β P (4) Probability of success (p) is same for all trials Let X : be the number of defective articles Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 β p = 1 β 1/10 = 9/10 Hence, P(X = x) = 12Cx (π/ππ)^π (π/ππ)^(ππβπ) Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 β p = 1 β 1/10 = 9/10 Hence, P(X = x) = 12Cx (π/ππ)^π (π/ππ)^(ππβπ) We need to find Probability of getting 9 defective articles out of 12 i.e. P(X = 9) P(X = 9) = 12C9 (1/10)^9 (9/10)^(12 β 9) = (12 !)/((12 β 9) ! 9 !) (1/10)^9 (9/10)^3 = (12 Γ 11 Γ 10 Γ 9 !)/(3 ! 9 !) \ Γ1/γ10γ^9 (9/10)^3 = (12 Γ 11 Γ 10)/(3 Γ 2 Γ 1) Γ 1/γ10γ^9 (9/10)^3= (2 Γ 11)/γ10γ^8 9^3/γ10γ^3 = 22 (π^π/γππγ^ππ )