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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, 8 Deleted for CBSE Board 2023 Exams
Ex 13.5, 9 Deleted for CBSE Board 2023 Exams
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
Ex 13.5, 13 Important Deleted for CBSE Board 2023 Exams
Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams You are here
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams
Last updated at March 30, 2023 by Teachoo
Ex 13.5, 14 In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is (A) 10–1 (B) (1/2)^5 (C) (9/10)^5 (D) 9/10aLet X : be the number of defective bulbs Picking bulbs is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of times we pick a bulb = 5 p = Probability of getting defective bulb = 10/100 = 1/10 q = 1 – p = 1 – 1/10 = 9/10 Hence, P(X = x) = 5Cx (𝟏/𝟏𝟎)^𝒙 (𝟗/𝟏𝟎)^(𝟒−𝒙) We need to find Probability that no bulb is defective i.e. P(X = 0) P(X = 0) = 5C0(1/10)^0 (9/10)^(5 −0) = 1 × 1 × (9/10)^5 = (9/10)^5 ∴ Option C is the correct answer