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

Transcript

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’’^(๐’โˆ’๐’™) ๐’‘^๐’™ Here n = number of bulbs picked = 5 p = Probability of getting fused bulb = 0.05 q = 1 โˆ’ p = 1 โˆ’ 0.05 = 0.95 Hence, P(X = x) = 5Cx (๐ŸŽ.๐ŸŽ๐Ÿ“)^๐’™ (๐ŸŽ.๐Ÿ—๐Ÿ“)^(๐Ÿ“โˆ’๐’™) Probability that out of 5 such bulbs none fuses i.e. P(X = 0) P(X = 0) = 5C0 (0.05)^0 (0.95)^(5โˆ’0) = 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 โ‰ค 1) = P(X = 0) + P(X = 1) = 5C0 (0.05)^0 (0.95)^(5โˆ’0) + 5C1 (0.05)^1 (0.95)^(5โˆ’1) = 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 โˆ’ P (not more than one) = 1 โˆ’ ((0.95)4 ร— 1.2) (iv) probability that out of 5 such bulbs atleast one fuses P (At least one) = 1 โˆ’ P (not even one fuses ) = 1 โˆ’ P (0) = 1 โˆ’ (0.95)5

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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.