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Ex 13.5, 5 - The probability that a bulb produced by a factory will

Ex 13.5, 5 - Chapter 13 Class 12 Probability - Part 2
Ex 13.5, 5 - Chapter 13 Class 12 Probability - Part 3
Ex 13.5, 5 - Chapter 13 Class 12 Probability - Part 4


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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 𝒒^(π’βˆ’π’™) 𝒑^𝒙 Here n = number of bulbs picked = 5 p = Probability of getting fused bulb = 0.05 q = 1 βˆ’ p = 1 βˆ’ 0.05 = 0.95 Hence, P(X = x) = 5Cx (𝟎.πŸŽπŸ“)^𝒙 (𝟎.πŸ—πŸ“)^(πŸ“βˆ’π’™) Probability that out of 5 such bulbs none fuses i.e. P(X = 0) P(X = 0) = 5C0 (0.05)^0 (0.95)^(5βˆ’0) = 1 Γ— 1 Γ— (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 ≀ 1) = P(X = 0) + P(X = 1) = 5C0 (0.05)^0 (0.95)^(5βˆ’0) + 5C1 (0.05)^1 (0.95)^(5βˆ’1) = 1 Γ— 1 Γ— (0.95)5 + 5 Γ— 0.05 Γ— (0.95)4 = (0.95)4 [0.95+0.25] = (0.95)4 Γ— 1.2 (iii) Probability that out of 5 such bulbs more than one fuses P (more than one) = 1 βˆ’ P (not more than one) = 1 βˆ’ ((0.95)4 Γ— 1.2) (iv) probability that out of 5 such bulbs atleast one fuses P (At least one) = 1 βˆ’ P (not even one fuses ) = 1 βˆ’ P (0) = 1 βˆ’ (0.95)5

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.