Check sibling questions

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

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


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 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here n = number of bulbs picked = 5 p = Probability of getting fused bulb = 0.05 q = 1 − p = 1 − 0.05 = 0.95 Hence, P(X = x) = 5Cx (𝟎.𝟎𝟓)^𝒙 (𝟎.𝟗𝟓)^(𝟓−𝒙) Probability that out of 5 such bulbs none fuses i.e. P(X = 0) P(X = 0) = 5C0 (0.05)^0 (0.95)^(5−0) = 1 × 1 × (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)^(5−0) + 5C1 (0.05)^1 (0.95)^(5−1) = 1 × 1 × (0.95)5 + 5 × 0.05 × (0.95)4 = (0.95)4 [0.95+0.25] = (0.95)4 × 1.2 (iii) Probability that out of 5 such bulbs more than one fuses P (more than one) = 1 − P (not more than one) = 1 − ((0.95)4 × 1.2) (iv) probability that out of 5 such bulbs atleast one fuses P (At least one) = 1 − P (not even one fuses ) = 1 − P (0) = 1 − (0.95)5

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.