Question

It is known that 3% of plastic buckets manufactured in a factory are defective. Using the Poisson distribution on a sample of 100 buckets, find the probability of:

(i) Zero defective buckets

(ii) At most one bucket is defective

[Use e⁻³ = 0.049]

Answer 1

Probability of defective bucket = 0.03

n = 100

m = np = 100 x 0.03 = 3

Let X = number of defective buckets in a sample of 100

P (X = r) = \(\frac{m^re^{-m}}{r!}\), r = 0,1,2,3, ….

(i) P (no defective bucket) = P(r = 0 ) = \(\frac{3^0e^{-m}}{0!}\) = 0.049

(ii) P (at most one defective bucket) = P(r = 0, 1)

\(=\frac{3^0e^{-m}}{0!}+ \frac{3^1e^{-3}}{1!}\)

= 0.049 + 0.147

= 0.196