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: \(e^{-7.5} = 0.000553\)]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.00415
The number of defective items in the packet is binomial with n = 250 and p = 0.03. Since n is large and p is small, it is approximated by a Poisson distribution (the given value of \(e^{-7.5}\) points to this) with mean
\(\lambda = np = 250 \times 0.03 = 7.5\)
\( \displaystyle P(X = r) = \frac{e^{-\lambda}\lambda^{r}}{r!} \)
\( \displaystyle P(X = 1) = \frac{e^{-7.5} \times 7.5}{1!} = 0.000553 \times 7.5 \)
\(= 0.0041475 \approx 0.00415\)