Ex 13.5, 5

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﷯ where 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