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Binomial: defective lenses in packet

Short answer

Binomial n=10, p=0.002: P(no defective)=0.9802; P(at least one)=0.0198; P(at most two)≈1.0000.

In certain factory turning out optical lenses, there is a small chance, 1/500 for any lens to be defective. The lenses are supplied in packets of 10 each. What is the probability that a packet will contain i) No defective lens ii) At least one defective lenses iii) At most two defective lenses

Each lens is defective with probability p = 1/500 = 0.002, independently. For a packet of n = 10, the number of defective lenses X ~ B(10, 0.002):

P(X = x) = C(10, x)(0.002)ˣ(0.998)¹⁰⁻ˣ

P(0) = 0.998¹⁰               = 0.98018
P(1) = 10 × 0.002 × 0.998⁹   = 0.01964
P(2) = 45 × 0.002² × 0.998⁸  = 0.00018

i) No defective lens

P(X = 0) = 0.9802

ii) At least one defective lens

P(X ≥ 1) = 1 − P(0) = 1 − 0.98018 = 0.0198

iii) At most two defective lenses

P(X ≤ 2) = 0.98018 + 0.01964 + 0.00018 = 0.99999 ≈ 1.0000

(Poisson approximation with λ = np = 0.02 gives almost the same values: 0.9802, 0.0198, 1.0000.)

Answer: (i) 0.9802, (ii) 0.0198, (iii) ≈ 1.0000

✓ Answer checked by Claude Sonnet 5.5

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