# Ex 13.4, 6

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 𝒒𝒏−𝒙 𝒑𝒙 n = number of times we pick a bulb = 4 p = Probability of getting defective bulb = 630 = 15 q = 1 – p = 1 – 15 = 45 Hence, P(X = x) = 4Cx 𝟏𝟓𝒙 𝟒𝟓𝟒−𝒙 Now, P(X = 0) = 4C0 150 454−0= 4C0 150 454 = 256625 P(X = 1) = 4C1 151 454−1= 4C1 151 453 = 4 × 64625 = 256625 P(X = 2) = 4C2 152 454−2= 4C2 152 452 = 6 × 16625 = 96625 P(X = 3) = 4C3 153 454−3= 4C3 153 451 = 4 × 4625 = 16625 P(X = 4) = 4C4 154 454−4= 4C4 154 450= 1625 So, the probability distribution is

Chapter 13 Class 12 Probability

Serial order wise

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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.