In a certain factory turning razor blades, there is a small chance $\frac{1}{500}$ for any blade to be defective. The blades are in packets of 10. Use Poisson's distribution to calculate the approximate number of packets containing no defective, one defective and two defective blades respectively in a consignment of 10000 packets. (Use $e^{-0.02} = 0.9802$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 9802, 196, and 2 packets
Here $N = 10000, n = 10, p =\frac{1}{500}=0.002$
$⇒λ= np = 10 × 0.002⇒ λ = 0.02$
P (No defective blade) = $P(X = 0) =\frac{e^{-0.02} (0.02)^0}{0!}$
$= e^{-0.02} = 0.9802$
∴ The approximate number of packets containing no defective blade
$= 10000 x 0.9802 = 9802$
P (one defective blade) = $P(X = 1) =\frac{e^{-0.02} (0.02)^1}{1!}$
$= 0.9802 × 0.02 0.0196$
∴ The approximate number of packets containing one defective blade = $10000 × 0.0196 = 196$
P (two defective blades) = $P(X = 2) =\frac{e^{-0.02} (0.02)^2}{2!}=\frac{0.9802 × 0.0004}{2}$
$= 0.000196 = 0.0002$
∴ The approximate number of packets containing two defective blades = $10000 × 0.0002 = 2$.