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Ex 13.4
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
Ex 13.4, 11 Important
Ex 13.4, 12 Important
Ex 13.4, 13 Important Deleted for CBSE Board 2023 Exams
Ex 13.4, 14
Ex 13.4, 15
Ex 13.4, 16 (MCQ)
Ex 13.4, 17 (MCQ) Important
Last updated at March 30, 2023 by Teachoo
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