Question

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 defective bulbs using Poisson distribution

Answer 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) =  $e^{-\lambda} \dfrac{{\lambda}^x}{x!}$

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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