Question

In a workshop of 100 machines, 25 machines are defective. Assuming the Poisson law for the number of defective machines, the probability that a random sample of 5 machines will have no defective machine is

• A. \( e^{-0.25} \)
• B. \( 0.25\, e^{-0.25} \)
• C. \( 0.25\, e^{-1.25} \)
• D. \( e^{-1.25} \)

Hint:

For small probability and large number of trials, use Poisson approximation with \( \lambda = n \cdot p \). The probability of zero occurrences is \( e^{-\lambda} \).

Answer 1

The Correct Option is D

Total machines = 100, defective = 25 So, probability of a machine being defective = \( \frac{25}{100} = 0.25 \) Let \( X \) be the number of defective machines in a sample of 5. Then \( X \sim \text{Poisson}(\lambda = 5 \times 0.25 = 1.25) \) We are required to find: \[ P(X = 0) = \frac{(1.25)^0}{0!} e^{-1.25} = 1 \cdot e^{-1.25} = e^{-1.25} \]