Question

In a component manufacturing industry, there is a small probability for any 

component to be defective. The components are supplied in packets of . Use 

Poisson approximation to calculate appropriately the number of packets containing 

 (i) at least one defective (ii) at most one defective in a consignment of 100 packets

Answer 1

The number of defective blades in a packet has binomial distribution 𝐵(𝑛, 𝑝) with

parameters 𝑛 = 100 and 𝑝 = 0.0002

The binomial distribution can be approximated using Poisson with parameter 𝑚 =

𝑛𝑝 = 0.02

Let 𝑋 equals the number of defective blades in a packet.

$𝑝_0 = 𝑃(𝑋 = 0) = 𝑒^{-0.02} = 0.9802$

Using the formula $𝑝_{𝑥+1 }= 𝑝𝑥 ⋅ \frac{𝑚 }{ 𝑥 + 1}$

we have:

$𝑝_1 = 𝑝_0 ⋅ \frac{ 0.02}{ 1} = 0.019604$

$𝑝_2 = 𝑝_1 ⋅\frac{ 0.02}{ 2} = 0.009802$

$𝑝_3 = 𝑝_2 ⋅ \frac{ 0.02}{ 3}≈ 0$

Thus expected frequencies are:

(i) at least one defective 

$𝑛_1+𝑛_2+𝑛_3 =100 ⋅ 𝑝_1+100 ⋅ 𝑝_2+100 ⋅ 𝑝_3 ≈ 1.96+0.9802+0 ≈ 2.9402$

ii) at most one defective

$𝑛_0+𝑛_1=100 ⋅ 𝑝_0+100 ⋅ 𝑝_1 ≈ 0.019604+0.009802 ≈ 0.029406$