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Last updated at Feb. 17, 2020 by Teachoo
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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 ๐^(๐โ๐) ๐^๐ n = number of times we pick a bulb = 4 p = Probability of getting defective bulb = 6/30 = 1/5 q = 1 โ p = 1 โ 1/5 = 4/5 Hence, P(X = x) = 4Cx (๐/๐)^๐ (๐/๐)^(๐โ๐) Now, P(X = 0) = 4C0 (1/5)^0 (4/5)^(4โ0)= 4C0 (1/5)^0 (4/5)^4 = 256/625 P(X = 1) = 4C1 (1/5)^1 (4/5)^(4โ1)= 4C1 (1/5)^1 (4/5)^3 = 4 ร 64/625 = 256/625 P(X = 2) = 4C2 (1/5)^2 (4/5)^(4โ2)= 4C2 (1/5)^2 (4/5)^2 = 6 ร 16/625 = 96/625 P(X = 3) = 4C3 (1/5)^3 (4/5)^(4โ3)= 4C3 (1/5)^3 (4/5)^1 = 4 ร 4/625 = 16/625 P(X = 4) = 4C4 (1/5)^4 (4/5)^(4โ4)= 4C4 (1/5)^4 (4/5)^0= 1/625 So, the probability distribution is
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Ex 13.4, 6 Important You are here
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Ex 13.4, 10 Not in Syllabus - CBSE Exams 2021
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