It is given that 3% of items manufactured by an industry are defective. The probability that a packet of 250 items contains one defective item is: [Given: $e^{-7.5}= 0.000553$]

0.00415

The correct answer is Option (3) → 0.00415

Defective probability: $p = 0.03$, Number of items: $n = 250$, Number of defective items: $x = 1$

Use Poisson approximation: $\lambda = n \cdot p = 250 \cdot 0.03 = 7.5$

Poisson probability: $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$

$P(X=1) = \frac{e^{-7.5} \cdot 7.5^1}{1} = 7.5 \cdot e^{-7.5}$

$P = 7.5 \cdot e^{-7.5} = 0.00415 $