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Cycle tyres are supplied in lots of 10 and there is a chance of 1 in 500 tyres to be defective. Using Poisson’s distribution the approximate number of lots containing no defective tyres in a consignment of 10,000 lots is

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9998

9982

9980

9802

answer is C.

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Detailed Solution

$λ = \frac{1}{500} \times 10 = 0.02$

$P (X = 0) = e^{- 0.02}$

$= 1 - \frac{0.02}{1!} + \frac{{(0.02)}^{2}}{2!} ........$

$= 0.9802$

The number of lots containing no defective tyre in

10000 lots = $10, 000 \times 0.9802 = 9802$

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