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  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

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’’^(๐’โˆ’๐’™) 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Ÿ/๐Ÿ“)^๐’™ (๐Ÿ’/๐Ÿ“)^(๐Ÿ’โˆ’๐’™) Now, P(X = 0) = 4C0 (1/5)^0 (4/5)^(4โˆ’0)= 4C0 (1/5)^0 (4/5)^4 = 256/625 P(X = 1) = 4C1 (1/5)^1 (4/5)^(4โˆ’1)= 4C1 (1/5)^1 (4/5)^3 = 4 ร— 64/625 = 256/625 P(X = 2) = 4C2 (1/5)^2 (4/5)^(4โˆ’2)= 4C2 (1/5)^2 (4/5)^2 = 6 ร— 16/625 = 96/625 P(X = 3) = 4C3 (1/5)^3 (4/5)^(4โˆ’3)= 4C3 (1/5)^3 (4/5)^1 = 4 ร— 4/625 = 16/625 P(X = 4) = 4C4 (1/5)^4 (4/5)^(4โˆ’4)= 4C4 (1/5)^4 (4/5)^0= 1/625 So, the probability distribution is

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.