
Binomial Distribution
Ex 13.5, 12 Deleted for CBSE Board 2022 Exams
Ex 13.5, 2 Deleted for CBSE Board 2022 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 9 Deleted for CBSE Board 2022 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 11 Deleted for CBSE Board 2022 Exams
Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams You are here
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
Example 32 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams
Misc 6 Important Deleted for CBSE Board 2022 Exams
Misc 7 Important
Ex 13.5, 7 Important Deleted for CBSE Board 2022 Exams
Example 31 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 5 Deleted for CBSE Board 2022 Exams
Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams
Misc 5 Important Deleted for CBSE Board 2022 Exams
Misc 9 Deleted for CBSE Board 2022 Exams
Misc 4 Deleted for CBSE Board 2022 Exams
Misc 10 Important
Example 35 Deleted for CBSE Board 2022 Exams
Ex 13.5, 8 Deleted for CBSE Board 2022 Exams
Example 34 Deleted for CBSE Board 2022 Exams
Binomial Distribution
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