in a certain factory turnning out razor blades,there is a small chance of 0.002 for any blade to be defective.the blades are supplied in packets of 10,use poisson distribution to calculate the approximate number of packets containg no defective,one defective and two defective blades respectively in a consignment of 10000 packets

# Finding probability of defective blades

**Harsha_1997**#4

The probability of a defect per blade is p = 1/500 = 0.002. This means that for a packet of 10, the mean

number of defects L = 10p = 0.02. The parameter L is used in the Poisson distribution to give the probability

of the number of defects, n, in a packet of 10:

𝑃(𝑛) =L^n/𝑛! *e^-L

𝑃(0)=0.02^0/0!*e^-0.022 ≈ 0.98;

The approximate number of packets containing blades with no defective blades is 𝑃(0) ∗ 10000 ≈ 9800

𝑃(1) =0.02^1/1!*e^-0.02≈ 0.0196;

The approximate number of packets containing blades with one defective blade is 𝑃(1) ∗ 10000 ≈ 196

𝑃(2) =0.02^2/2!*e^-0.022 ≈ 0.000196;

The approximate number of packets containing blades with two defective blades is 𝑃(2) ∗ 10000 ≈ 2

𝑃(3) =0.02^3/3!*e^-0.022 ≈ 0.0000013;

The approximate number of packets containing blades with three defective blades is 𝑃(3) ∗ 10000 ≈ 0

therefore sollution is: 9800; 196; 2; 0.