If 2 per cent bulbs are known to be defective bulbs, find the probability that in a lot of 300 bulbs, there will be 2 or 3 defecti

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1. Given : 2 per cent bulbs are known to be defective bulbs,

   To Find : the probability that in a lot of 300 bulbs, there will be 2 or 3 defective bulbs using Poisson distribution

   Solution:

   Binomial distribution

   P (x) = ⁿCₓpˣqⁿ⁻ˣ q = 1 – p

   Poisson approximation for large n

   P(x) = [tex]e^{-\lambda} \dfrac{{\lambda}^x}{x!}[/tex]

   where λ = np = constant

   Here λ = 300(0.02) = 6

   P(2) = e⁻⁶ . 6²/2! = 0.044618

   P(3) = e⁻⁶ . 6³/3! = 0.089235

   probability that in a lot of 300 bulbs, there will be 2 or 3 defective bulbs

   = 0.044618 + 0.089235

   = 0.133853

   probability that in a lot of 300 bulbs, there will be 2 or 3 defective bulbs using Poisson distribution = 0.133853

   Additional Info:

   Using normal Distribution

   P(2) = ³⁰⁰C₂(0.02)²(0.98)²⁹⁸ = 0.0435705

   P(3) = ³⁰⁰C₃(0.02)³(0.98)²⁹⁷ = 0.0883267

   probability that in a lot of 300 bulbs, there will be 2 or 3 defective bulbs

   = 0.1318972

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