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Ex 13.5
Ex 13.5, 2 Deleted for CBSE Board 2023 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2023 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2023 Exams
Ex 13.5, 5 Deleted for CBSE Board 2023 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2023 Exams
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Ex 13.5, 10 Important Deleted for CBSE Board 2023 Exams
Ex 13.5, 11 Deleted for CBSE Board 2023 Exams
Ex 13.5, 12 Deleted for CBSE Board 2023 Exams
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Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 by Teachoo
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 – 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 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 – p = 1 – 1/10 = 9/10 Hence, P(X = x) = 12Cx (𝟏/𝟏𝟎)^𝒙 (𝟗/𝟏𝟎)^(𝟏𝟐−𝒙) Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 – p = 1 – 1/10 = 9/10 Hence, P(X = x) = 12Cx (𝟏/𝟏𝟎)^𝒙 (𝟗/𝟏𝟎)^(𝟏𝟐−𝒙) 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 − 9) = (12 !)/((12 − 9) ! 9 !) (1/10)^9 (9/10)^3 = (12 × 11 × 10 × 9 !)/(3 ! 9 !) \ ×1/〖10〗^9 (9/10)^3 = (12 × 11 × 10)/(3 × 2 × 1) × 1/〖10〗^9 (9/10)^3= (2 × 11)/〖10〗^8 9^3/〖10〗^3 = 22 (𝟗^𝟑/〖𝟏𝟎〗^𝟏𝟏 )