Question

In a factory turning out razor blade, there is a small chance of 1/500 for any blade to be defective. The blades are supplied in a packet of 10. Use Poisson distribution to calculate the approximate number of packets containing blades with no defective, one defective, two defectives and three defectives in a consignment of 10,000 packets

Answer 1

Answer on Question #39618 – Math – Other

In a factory turning out razor blade, there is a small chance of 1/500 for any blade to be defective. The blades are supplied in a packet of 10. Use Poisson distribution to calculate the approximate number of packets containing blades with no defective, one defective, two defectives and three defectives blades in a consignment of 10,000 packets.

Solution:

From the conditions above, the probability of defect per razor blade is $p = \frac{1}{500} = 0,002$.

$$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} - \text{Poisson distribution, where } k \in \mathbb{N} \text{ and } \lambda = 10 \cdot 0,002 = 0,02,$$ $$k = 0,1,2,3, \text{ k - razors number of defects};$$ $$e^{-\lambda} = e^{-0,02} \approx 2,7^{-0,02} = 0,9803;$$ $$P(X = 0) = \frac{1 \cdot 0,9803}{0!} = 0,9803,$$ $$P(X = 1) = \frac{0.02^1 \cdot 0,9803}{1!} = 0,0196,$$ $$P(X = 2) = \frac{0.02^2 \cdot 0.9803}{2!} = 0,000196,$$ $$P(X = 3) = \frac{0.02^3 \cdot 0.9803}{3!} \approx 0,0000013.$$

Then let calculate the approximate number of packets containing blades with 0,1,2,3 defects $n_{k=0} = 10000 \cdot 0,9803 = 9803$,

$$n_{k=1} = 10000 \cdot 0,0196 = 196,$$ $$n_{k=2} = 10000 \cdot 0,000196 = 2,$$ $$n_{k=0} = 10000 \cdot 0,0000013 \approx 0,01 = 0.$$

Answer: $n_{k=0} = 9803$, $n_{k=1} = 196$, $n_{k=2} = 2$, $n_{k=0} = 0$.