Solve all your doubts with Teachoo Black (new monthly pack available now!)

Ex 13.5

Ex 13.5, 1
Deleted for CBSE Board 2023 Exams

Ex 13.5, 2 Deleted for CBSE Board 2023 Exams

Ex 13.5, 3 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 4 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 5 Deleted for CBSE Board 2023 Exams You are here

Ex 13.5, 6 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 7 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 8 Deleted for CBSE Board 2023 Exams

Ex 13.5, 9 Deleted for CBSE Board 2023 Exams

Ex 13.5, 10 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 11 Deleted for CBSE Board 2023 Exams

Ex 13.5, 12 Deleted for CBSE Board 2023 Exams

Ex 13.5, 13 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at Feb. 17, 2020 by Teachoo

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 π^(πβπ) π^π Here n = number of bulbs picked = 5 p = Probability of getting fused bulb = 0.05 q = 1 β p = 1 β 0.05 = 0.95 Hence, P(X = x) = 5Cx (π.ππ)^π (π.ππ)^(πβπ) Probability that out of 5 such bulbs none fuses i.e. P(X = 0) P(X = 0) = 5C0 (0.05)^0 (0.95)^(5β0) = 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 β€ 1) = P(X = 0) + P(X = 1) = 5C0 (0.05)^0 (0.95)^(5β0) + 5C1 (0.05)^1 (0.95)^(5β1) = 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 β P (not more than one) = 1 β ((0.95)4 Γ 1.2) (iv) probability that out of 5 such bulbs atleast one fuses P (At least one) = 1 β P (not even one fuses ) = 1 β P (0) = 1 β (0.95)5