# Ex 13.4, 6 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 17, 2020 by

Last updated at Feb. 17, 2020 by

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Ex 13.4, 6 From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.Let 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 n = number of times we pick a bulb = 4 p = Probability of getting defective bulb = 6/30 = 1/5 q = 1 – p = 1 – 1/5 = 4/5 Hence, P(X = x) = 4Cx (𝟏/𝟓)^𝒙 (𝟒/𝟓)^(𝟒−𝒙) Now, P(X = 0) = 4C0 (1/5)^0 (4/5)^(4−0)= 4C0 (1/5)^0 (4/5)^4 = 256/625 P(X = 1) = 4C1 (1/5)^1 (4/5)^(4−1)= 4C1 (1/5)^1 (4/5)^3 = 4 × 64/625 = 256/625 P(X = 2) = 4C2 (1/5)^2 (4/5)^(4−2)= 4C2 (1/5)^2 (4/5)^2 = 6 × 16/625 = 96/625 P(X = 3) = 4C3 (1/5)^3 (4/5)^(4−3)= 4C3 (1/5)^3 (4/5)^1 = 4 × 4/625 = 16/625 P(X = 4) = 4C4 (1/5)^4 (4/5)^(4−4)= 4C4 (1/5)^4 (4/5)^0= 1/625 So, the probability distribution is

Ex 13.4

Ex 13.4, 1

Ex 13.4, 2

Ex 13.4, 3 Important

Ex 13.4, 4 (i)

Ex 13.4, 4 (ii)

Ex 13.4, 4 (iii) Important

Ex 13.4, 5 (i) Important

Ex 13.4, 5 (ii)

Ex 13.4, 6 Important You are here

Ex 13.4, 7 Important

Ex 13.4, 8

Ex 13.4, 9

Ex 13.4, 10 Deleted for CBSE Board 2022 Exams

Ex 13.4, 11 Important Deleted for CBSE Board 2022 Exams

Ex 13.4, 12 Important Deleted for CBSE Board 2022 Exams

Ex 13.4, 13 Important Deleted for CBSE Board 2022 Exams

Ex 13.4, 14 Deleted for CBSE Board 2022 Exams

Ex 13.4, 15 Deleted for CBSE Board 2022 Exams

Ex 13.4, 16 (MCQ) Deleted for CBSE Board 2022 Exams

Ex 13.4, 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Chapter 13 Class 12 Probability (Term 2)

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