# Ex 13.5, 5 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 17, 2020 by Teachoo

Last updated at Feb. 17, 2020 by Teachoo

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Ex 13.5, 5 The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use.Let X : Number of bulbs fused Picking a bulb is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx ๐^(๐โ๐) ๐^๐ Here n = number of bulbs picked = 5 p = Probability of getting fused bulb = 0.05 q = 1 โ p = 1 โ 0.05 = 0.95 Hence, P(X = x) = 5Cx (๐.๐๐)^๐ (๐.๐๐)^(๐โ๐) Probability that out of 5 such bulbs none fuses i.e. P(X = 0) P(X = 0) = 5C0 (0.05)^0 (0.95)^(5โ0) = 1 ร 1 ร (0.95)5 = (0.95)5 (ii) Probability that out of 5 such bulbs not more than one fuses P(not not than one) = P( X โค 1) = P(X = 0) + P(X = 1) = 5C0 (0.05)^0 (0.95)^(5โ0) + 5C1 (0.05)^1 (0.95)^(5โ1) = 1 ร 1 ร (0.95)5 + 5 ร 0.05 ร (0.95)4 = (0.95)4 [0.95+0.25] = (0.95)4 ร 1.2 (iii) Probability that out of 5 such bulbs more than one fuses P (more than one) = 1 โ P (not more than one) = 1 โ ((0.95)4 ร 1.2) (iv) probability that out of 5 such bulbs atleast one fuses P (At least one) = 1 โ P (not even one fuses ) = 1 โ P (0) = 1 โ (0.95)5

Ex 13.5

Ex 13.5, 1
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Chapter 13 Class 12 Probability (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.